Resolve as seguintes equações usando a fórmula resolvente

Fórmula resolvente da equação do 2.º grau:

$$\begin{array}{*{20}{c}}
{a{x^2} + bx + c = 0}& \Leftrightarrow &{x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}}&{(a \ne 0)}
\end{array}$$

1. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {2{x^2} + 4x – 4 = 0}& \Leftrightarrow &{{x^2} + 2x – 2 = 0} \\
   {}& \Leftrightarrow &{x = \frac{{ – 2 \pm \sqrt {{2^2} – 4 \times 1 \times ( – 2)} }}{{2 \times 1}}} \\
   {}& \Leftrightarrow &{x = \frac{{ – 2 \pm \sqrt {12} }}{2}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = \frac{{ – 2 – \sqrt {12} }}{2}}& \vee &{x = \frac{{ – 2 + \sqrt {12} }}{2}}
   \end{array}}
   \end{array}$$
2. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {6{x^2} + 5x + 1 = 0}& \Leftrightarrow &{x = \frac{{ – 5 \pm \sqrt {{{( – 5)}^2} – 4 \times 6 \times 1} }}{{2 \times 6}}} \\
   {}& \Leftrightarrow &{x = \frac{{ – 5 \pm \sqrt 1 }}{{2 \times 6}}} \\
   {}& \Leftrightarrow &{x = \frac{{ – 5 \pm 1}}{{12}}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x =  – \frac{1}{2}}& \vee &{x =  – \frac{1}{3}}
   \end{array}}
   \end{array}$$
3. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {{x^2} – 4x + 4 = 0}& \Leftrightarrow &{x = \frac{{4 \pm \sqrt {{{( – 4)}^2} – 4 \times 1 \times 4} }}{{2 \times 1}}} \\
   {}& \Leftrightarrow &{x = \frac{{4 \pm \sqrt 0 }}{2}} \\
   {}& \Leftrightarrow &{x = \frac{{4 \pm 0}}{2}} \\
   {}& \Leftrightarrow &{x = 2}
   \end{array}$$
4. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {{x^2} – 3x + 2 = 0}& \Leftrightarrow &{x = \frac{{3 \pm \sqrt {{{( – 3)}^2} – 4 \times 1 \times 2} }}{{2 \times 1}}} \\
   {}& \Leftrightarrow &{x = \frac{{3 \pm \sqrt 1 }}{2}} \\
   {}& \Leftrightarrow &{x = \frac{{3 \pm 1}}{2}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 1}& \vee &{x = 2}
   \end{array}}
   \end{array}$$
5. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {{x^2} – \frac{5}{3}x – \frac{2}{3} = 0}& \Leftrightarrow &{3{x^2} – 5x – 2 = 0} \\
   {}& \Leftrightarrow &{x = \frac{{5 \pm \sqrt {{{( – 5)}^2} – 4 \times 3 \times ( – 2)} }}{{2 \times 3}}} \\
   {}& \Leftrightarrow &{x = \frac{{5 \pm \sqrt {49} }}{6}} \\
   {}& \Leftrightarrow &{x = \frac{{5 \pm 7}}{6}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x =  – \frac{1}{3}}& \vee &{x = 2}
   \end{array}}
   \end{array}$$
6. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {x\left( {x – 8} \right) =  – 42 + 5x}& \Leftrightarrow &{{x^2} – 8x – 5x + 42 = 0} \\
   {}& \Leftrightarrow &{{x^2} – 13x + 42 = 0} \\
   {}& \Leftrightarrow &{x = \frac{{13 \pm \sqrt {{{( – 13)}^2} – 4 \times 1 \times 42} }}{{2 \times 1}}} \\
   {}& \Leftrightarrow &{x = \frac{{13 \pm \sqrt 1 }}{2}} \\
   {}& \Leftrightarrow &{x = \frac{{13 \pm 1}}{2}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 6}& \vee &{x = 7}
   \end{array}}
   \end{array}$$
7. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {4x\left( {2x – 5} \right) = 3x – 14}& \Leftrightarrow &{8{x^2} – 20x – 3x + 14 = 0} \\
   {}& \Leftrightarrow &{8{x^2} – 23x + 14 = 0} \\
   {}& \Leftrightarrow &{x = \frac{{23 \pm \sqrt {{{( – 23)}^2} – 4 \times 8 \times 14} }}{{2 \times 8}}} \\
   {}& \Leftrightarrow &{x = \frac{{23 \pm \sqrt {81} }}{{16}}} \\
   {}& \Leftrightarrow &{x = \frac{{23 \pm 9}}{{16}}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = \frac{7}{8}}& \vee &{x = 2}
   \end{array}}
   \end{array}$$
8. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {\frac{x}{{\mathop 4\limits_{(1)} }} – \frac{{{{\left( {x – 1} \right)}^2}}}{{\mathop 2\limits_{(2)} }} = \mathop 0\limits_{(4)} }& \Leftrightarrow &{x – 2{{\left( {x – 1} \right)}^2} = 0} \\
   {}& \Leftrightarrow &{x – 2\left( {{x^2} – 2x + 1} \right) = 0} \\
   {}& \Leftrightarrow &{ – 2{x^2} + 5x – 2 = 0} \\
   {}& \Leftrightarrow &{x = \frac{{ – 5 \pm \sqrt {{5^2} – 4 \times ( – 2) \times ( – 2)} }}{{2 \times ( – 2)}}} \\
   {}& \Leftrightarrow &{x = \frac{{ – 5 \pm \sqrt 9 }}{{ – 4}}} \\
   {}& \Leftrightarrow &{x = \frac{{ – 5 \pm 3}}{{ – 4}}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 2}& \vee &{x = \frac{1}{2}}
   \end{array}}
   \end{array}$$
9. Resolvendo a equação, temos:
   $$\begin{array}{*{20}{l}}
   {\mathop 5\limits_{(3)} \left( {3 + x} \right) = \frac{1}{{\mathop 3\limits_{(1)} }}{{\left( { – 3 – x} \right)}^2}}& \Leftrightarrow &{15\left( {3 + x} \right) = {{\left( { – 3 – x} \right)}^2}} \\
   {}& \Leftrightarrow &{45 + 15x = 9 + 6x + {x^2}} \\
   {}& \Leftrightarrow &{{x^2} – 9x – 36 = 0} \\
   {}& \Leftrightarrow &{x = \frac{{9 \pm \sqrt {{{( – 9)}^2} – 4 \times 1 \times ( – 36)} }}{{2 \times 1}}} \\
   {}& \Leftrightarrow &{x = \frac{{9 \pm \sqrt {225} }}{2}} \\
   {}& \Leftrightarrow &{x = \frac{{9 \pm 15}}{2}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x =  – 3}& \vee &{x = 12}
   \end{array}}
   \end{array}$$