Quadratic equation $4x^2+4x=7$ using quadratic formula Solve using quadratic formula. $4x^2+4x=7$
So $4x^2+4x-7=0$
$A=4$ $b=4$ $c=-7$
$$x=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}=\frac{-4\pm\sqrt{16+112}}{8}=\frac{-4\pm\sqrt{128}}{8}$$
What's next?

I can't see what you have written but it should look like:
$$\begin{align}
4x^2+4x-7&=0
\\ \Rightarrow x_{\pm}&=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}
\\ &=\frac{-4\pm{\sqrt{16+112}}}{8}
\\&=\frac{-4\pm\sqrt{128}}{8}
\\ &=\frac{-4+\sqrt{128}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{128}}{8}
\\ &=\frac{-4+\sqrt{8^2\cdot 2}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{8^2\cdot 2}}{8}
\\ &=\frac{-4+\sqrt{8^2}\cdot \sqrt{2}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{8^2}\cdot \sqrt{2}}{8}
\\ &=\frac{-4+8\cdot \sqrt{2}}{8}\,\,\,\text{ or }\frac{-4-8\cdot \sqrt{2}}{8}
\\ &=-\frac{1}{2}+\sqrt{2}\,\,\,\text{ or }-\frac{1}{2}-\sqrt{2}.
\end{align}$$
The approximate values of these may be found using a calculator.
