# How do you solve (-2x+5)^2 = -8?

Jul 29, 2017

See a solution process below:

#### Explanation:

First, expand the term on the left using this special rule for multiplying quadratics:

${\left(\textcolor{red}{x} + \textcolor{b l u e}{y}\right)}^{2} = {\textcolor{red}{x}}^{2} + 2 \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{y}}^{2}$

Substituting gives:

${\left(\textcolor{red}{- 2 x} + \textcolor{b l u e}{5}\right)}^{2} = - 8$

${\left(\textcolor{red}{- 2 x}\right)}^{2} + \left(2 \cdot \textcolor{red}{- 2 x} \cdot \textcolor{b l u e}{5}\right) + {\textcolor{b l u e}{5}}^{2} = - 8$

$4 {x}^{2} + \left(- 20 x\right) + 25 = - 8$

$4 {x}^{2} - 20 x + 25 = - 8$

We can next convert this to standard form:

$4 {x}^{2} - 20 x + 25 + \textcolor{red}{8} = - 8 + \textcolor{red}{8}$

$4 {x}^{2} - 20 x + 33 = 0$

We can now use the quadratic formula to find the solutions for $x$. The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{4}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 20}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{33}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \left(\textcolor{b l u e}{- 20}\right) \pm \sqrt{{\textcolor{b l u e}{- 20}}^{2} - \left(4 \cdot \textcolor{red}{4} \cdot \textcolor{g r e e n}{33}\right)}}{2 \cdot \textcolor{red}{4}}$

$x = \frac{\textcolor{b l u e}{20} \pm \sqrt{400 - 528}}{8}$

$x = \frac{\textcolor{b l u e}{20} \pm \sqrt{- 128}}{8}$

$x = \frac{\textcolor{b l u e}{20} \pm \sqrt{64 \cdot - 2}}{8}$

$x = \frac{\textcolor{b l u e}{20} \pm \sqrt{64} \sqrt{- 2}}{8}$

$x = \frac{\textcolor{b l u e}{20} \pm 8 \sqrt{- 2}}{8}$

Or

$x = \frac{\textcolor{b l u e}{20}}{8} \pm \frac{8 \sqrt{- 2}}{8}$

$x = \frac{5}{2} \pm \sqrt{- 2}$

Jul 29, 2017

No Real solutions;
within Complex numbers: $x = \frac{2}{5} - \sqrt{2} i \text{ or } \frac{2}{5} + \sqrt{2} i$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{xxx}} {\left(- 2 x + 5\right)}^{2} = - 8$

We note that any Real value squared must be $\ge 0$
therefore this equation has No valid Real solutions

If we are dealing with Complex values, then
$\textcolor{w h i t e}{\text{xxx}} {\left(- 2 x + 5\right)}^{2} = - 8$

$\textcolor{w h i t e}{\text{xxx}} 4 {x}^{2} - 20 x + 25 = - 8$

$\textcolor{w h i t e}{\text{xxx}} 4 {x}^{2} - 20 x + 33 = 0$

Then applying the quadratic formula that tells us that an equation of the form: $a {x}^{2} + b x + c = 0$
has solutions:
$\textcolor{w h i t e}{\text{xxx}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

We have solutions:
$\textcolor{w h i t e}{\text{xxx}} x = \frac{20 \pm \sqrt{400 - 4 \cdot 4 \cdot 33}}{2 \cdot 4}$

$\textcolor{w h i t e}{\text{xxx}} = \frac{20 \pm \sqrt{- 128}}{8}$

$\textcolor{w h i t e}{\text{xxx}} = \frac{20 \pm 8 \sqrt{2} i}{8}$

$\textcolor{w h i t e}{\text{xxx}} = \frac{5}{2} \pm \sqrt{2} i$