Quadratic equation $4x^2+4x=7$ using quadratic formula

Question

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?

Answer 1

Next is simplifying:

$$x=\frac{-4\pm\sqrt{4^2-4\cdot4\cdot-7}}{2\cdot4}=\frac{-4\pm\sqrt{16+112}}{8}=-\frac12\pm\frac18\cdot8\sqrt{2}=-\frac12\pm\sqrt2$$

Answer 2

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.