# How do you solve 2a^2 - 9a = 5?

Mar 22, 2016

$a = - \frac{1}{2} \text{ or } a = - 5$

#### Explanation:

Given:$\text{ } 2 {a}^{2} - 9 a = 5$

Write as:$\text{ } 2 \left({a}^{2} - \frac{9}{2} a - \frac{5}{2}\right) = 0$

Divide both sides by 2

${a}^{2} - \frac{9}{2} a - \frac{5}{2} = 0$

As there is no coefficient in front of ${a}^{2}$ there are two possible starting point

(-a+?)(-a+?)" or "(a+?)(a+?)

Lets investigate positive $a$

Notice that there are halves in this equation and we have a constant of $\frac{5}{2}$

Lets test $\frac{1}{2} \times 5$ giving

$\left(a + \frac{1}{2}\right) \left(a - 5\right) \to {a}^{2} + \frac{1}{2} a - 5 a - \frac{5}{2}$

But $\frac{1}{2} a - 5 a \to \frac{1}{2} a - \frac{10}{2} a = - \frac{9}{2} a \text{ }$ Which is what we need.
$\textcolor{b l u e}{\left(a + \frac{1}{2}\right) \left(a - 5\right) \to a = - \frac{1}{2} \text{ or } a = - 5}$