# How do you solve -4x^2+8x= -3?

Apr 18, 2018

$x = 1 \pm \frac{1}{2} \sqrt{7}$

#### Explanation:

$\text{rearrange the equation in "color(blue)"standard form}$

•color(white)(x)ax^2+bx+c=0

$\Rightarrow - 4 {x}^{2} + 8 x + 3 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{consider the a-c method of factorising}$

$\text{there are no whole numbers of - 12 which sum to + 8}$

$\text{we can solve using the method of "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

$\text{factor out } - 4$

$\Rightarrow - 4 \left({x}^{2} - 2 x - \frac{3}{4}\right) = 0$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - 2 x$

$\Rightarrow - 4 \left({x}^{2} + 2 \left(- 1\right) x \textcolor{red}{+ 1} \textcolor{red}{- 1} - \frac{3}{4}\right) = 0$

$\Rightarrow - 4 {\left(x - 1\right)}^{2} - 4 \left(- 1 - \frac{3}{4}\right) = 0$

$\Rightarrow - 4 {\left(x - 1\right)}^{2} + 7 = 0$

$\text{subtract 7 and divide by } - 4$

$\Rightarrow {\left(x - 1\right)}^{2} = \frac{7}{4}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow x - 1 = \pm \sqrt{\frac{7}{4}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\text{add 1 to both sides}$

$\Rightarrow x = 1 \pm \frac{1}{2} \sqrt{7} \leftarrow \textcolor{red}{\text{exact solutions}}$