# How do you solve using the completing the square method -3x^2-5x+5 = 33 ?

Mar 4, 2017

$x = - \frac{5}{6} \pm \frac{\sqrt{311}}{6} i$

#### Explanation:

Given:

$- 3 {x}^{2} - 5 x + 5 = 33$

Adding $3 {x}^{2} + 5 x - 5$ to both sides and transposing, we get:

$3 {x}^{2} + 5 x + 28 = 0$

This is in standard form:

$a {x}^{2} + b x + c = 0$

with $a = 3$, $b = 5$ and $c = 28$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {5}^{2} - 4 \left(3\right) \left(28\right) = 25 - 336 = - 311$

Since $\Delta < 0$, this quadratic has no Real solutions.

We can find Complex solutions by completing the square, then using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = \left(6 x + 5\right)$ and $B = \sqrt{311}$ as follows:

$0 = 12 \left(3 {x}^{2} + 5 x + 28\right)$

$\textcolor{w h i t e}{0} = 36 {x}^{2} + 60 x + 336$

$\textcolor{w h i t e}{0} = {\left(6 x\right)}^{2} + 2 \left(6 x\right) \left(5\right) + {5}^{2} + 311$

$\textcolor{w h i t e}{0} = {\left(6 x + 5\right)}^{2} + {\left(\sqrt{311}\right)}^{2}$

$\textcolor{w h i t e}{0} = {\left(6 x + 5\right)}^{2} - {\left(\sqrt{311} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(6 x + 5\right) - \sqrt{311} i\right) \left(\left(6 x + 5\right) + \sqrt{311} i\right)$

$\textcolor{w h i t e}{0} = \left(6 x + 5 - \sqrt{311} i\right) \left(6 x + 5 + \sqrt{311} i\right)$

Hence:

$x = \frac{1}{6} \left(- 5 \pm \sqrt{311} i\right) = - \frac{5}{6} \pm \frac{\sqrt{311}}{6} i$