# How do you solve for x in 2ax(x-3b)=9-(5a-6b)?

##### 1 Answer
Feb 13, 2018

See a solution process below:

#### Explanation:

First, expand the terms within parenthesis on both sides of the equation. Be careful to manage the signs correcly:

$\left(2 a x \cdot x\right) - \left(2 a x \cdot 3 b\right) = 9 - 5 a + 6 b$

$2 a {x}^{2} - 6 a b x = 9 - 5 a + 6 b$

Next, put the equation in standard form:

$2 a {x}^{2} - 6 a b x - \textcolor{red}{9} + \textcolor{b l u e}{5 a} - \textcolor{g r e e n}{6 b} = 9 - \textcolor{red}{9} - 5 a + \textcolor{b l u e}{5 a} + 6 b - \textcolor{g r e e n}{6 b}$

$2 a {x}^{2} - 6 a b x - 9 + 5 a - 6 b = 0 - 0 + 0$

$2 a {x}^{2} - 6 a b x + \left(- 9 + 5 a - 6 b\right) = 0$

We can use the quadratic equation to solve this problem 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:

$\textcolor{red}{2 a}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 6 a b}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 9 + 5 a - 6 b}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\left(- 6 a b\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 6 a b\right)}}^{2} - \left(4 \cdot \textcolor{red}{2 a} \cdot \textcolor{g r e e n}{\left(- 9 + 5 a - 6 b\right)}\right)}}{2 \cdot \textcolor{red}{2 a}}$

$x = \frac{\textcolor{b l u e}{6 a b} \pm \sqrt{36 {a}^{2} {b}^{2} - \left(8 a \textcolor{g r e e n}{\left(- 9 + 5 a - 6 b\right)}\right)}}{4 a}$

$x = \frac{\textcolor{b l u e}{6 a b} \pm \sqrt{36 {a}^{2} {b}^{2} - \left(- 72 a + 40 {a}^{2} - 48 a b\right)}}{4 a}$

$x = \frac{\textcolor{b l u e}{6 a b} \pm \sqrt{36 {a}^{2} {b}^{2} + 72 a - 40 {a}^{2} + 48 a b}}{4 a}$

$x = \frac{\textcolor{b l u e}{6 a b}}{4 a} \pm \frac{\sqrt{4 \left(9 {a}^{2} {b}^{2} + 18 a - 10 {a}^{2} + 12 a b\right)}}{4 a}$

$x = \frac{3 b}{2} \pm \frac{\sqrt{4} \sqrt{\left(9 {a}^{2} {b}^{2} + 18 a - 10 {a}^{2} + 12 a b\right)}}{4 a}$

$x = \frac{3 b}{2} \pm \frac{2 \sqrt{\left(9 {a}^{2} {b}^{2} + 18 a - 10 {a}^{2} + 12 a b\right)}}{4 a}$

$x = \frac{3 b}{2} \pm \frac{\sqrt{\left(9 {a}^{2} {b}^{2} + 18 a - 10 {a}^{2} + 12 a b\right)}}{2 a}$