# How do you solve 2x^2+9x+10=0?

Dec 22, 2016

$- \frac{5}{2} \mathmr{and} - 2$

#### Explanation:

Use improved quadratic formula (Socratic Search)
$y = 2 {x}^{2} + 9 x + 10 = 0$
Both roots are negative. (rule of signs)
$D = {d}^{2} = {b}^{2} - 4 a c = 81 - 80 = 1$ --> $d = \pm 1$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{9}{4} \pm \frac{1}{4} = \frac{- 9 \pm 1}{4}$
$x 1 = - \frac{10}{4} = - \frac{5}{2}$ and $x 2 = - \frac{8}{4} = - 2$

Dec 22, 2016

$x = - 2 \text{ }$ or $\text{ } x = - \frac{5}{2}$

#### Explanation:

Given:

$2 {x}^{2} + 9 x + 10 = 0$

We can use an AC method.

Find a pair of factors of $A C = 2 \cdot 10 = 20$ with sum $B = 9$

The pair $5 , 4$ works.

Use this pair to split the middle term and factor by grouping:

$0 = 2 {x}^{2} + 9 x + 10$

$\textcolor{w h i t e}{0} = 2 {x}^{2} + 5 x + 4 x + 10$

$\textcolor{w h i t e}{0} = \left(2 {x}^{2} + 5 x\right) + \left(4 x + 10\right)$

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

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

Hence solutions:

$x = - 2 \text{ }$ or $\text{ } x = - \frac{5}{2}$