# How do you solve  4x^2 + 23x + 15 = 0?

May 8, 2016

Factorise, then let each factor in turn be equal to 0.

$x = - \frac{3}{4} \mathmr{and} x = - 5$

#### Explanation:

$4 {x}^{2} + 23 x + 15 = 0 \text{ }$ Find factors of 4 and 15 which add to 23

$\left(4 x + 3\right) \left(x + 5\right) = 0 \text{ }$ 4x5 + 1x3 = 23

if $4 x + 3 = 0 \Rightarrow x = - \frac{3}{4}$
if $x + 5 = 0 \Rightarrow x = - 5$

May 8, 2016

Use the quadratic formula to find:

$x = - 5$ or $x = - \frac{3}{4}$

#### Explanation:

Use the quadratic formula.

The equation:

$4 {x}^{2} + 23 x + 15 = 0$

is in the form $a {x}^{2} + b x + c = 0$, with $a = 4$, $b = 23$ and $c = 15$.

This has roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 23 \pm \sqrt{{23}^{2} - \left(4 \cdot 4 \cdot 15\right)}}{2 \cdot 4}$

$= \frac{- 23 \pm \sqrt{529 - 240}}{8}$

$= \frac{- 23 \pm \sqrt{289}}{8}$

$= \frac{- 23 \pm 17}{8}$

That is:

$x = \frac{- 40}{8} = - 5$

or:

$x = \frac{- 6}{8} = - \frac{3}{4}$