# How do you solve 3x^2+14x+15=0?

Apr 20, 2018

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

#### Explanation:

$\implies 3 {x}^{2} + 14 x + 15 = 0$

It’s in the form of $a {x}^{2} + b x + c = 0$

where,

• $a = 3$
• $b = 14$
• $c = 15$

Use formula for quadratic equation to find $x$

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

x = (-14 +- sqrt(14^2 - (4 × 3 × 15)))/(2 × 3)

$x = \frac{- 14 \pm \sqrt{196 - 180}}{6}$

$x = \frac{- 14 \pm \sqrt{16}}{6}$

$x = \frac{- 14 \pm 4}{6}$

$x = \frac{- 14 + 4}{6} \textcolor{w h i t e}{\ldots .} \text{or} \textcolor{w h i t e}{\ldots .} x = \frac{- 14 - 4}{6}$

$x = \frac{- 10}{6} \textcolor{w h i t e}{\ldots \ldots \ldots .} \text{or} \textcolor{w h i t e}{\ldots .} x = \frac{- 18}{6}$

$x = - \frac{5}{3} \textcolor{w h i t e}{\ldots \ldots \ldots .} \text{or} \textcolor{w h i t e}{\ldots .} x = - 3$

Apr 20, 2018

I would use the quadratic formula.

#### Explanation:

The quadratic formula is applicable on equations of the form:

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

And looks like this:

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

In our case:

$a = 3$
$b = 14$
$c = 15$

$x = \frac{- 14 \pm \sqrt{{\left(14\right)}^{2} - 4 \left(3\right) \left(15\right)}}{\left(2\right) \left(3\right)}$

$x = \frac{- 14 \pm \sqrt{196 - 180}}{6}$

$x = \frac{- 14 \pm \sqrt{16}}{6}$

$x = \frac{- 14 \pm 4}{6}$

So:

${x}_{1} = \left(- \frac{18}{6}\right) = - 3$

${x}_{2} = \left(- \frac{10}{6}\right) = - \frac{5}{3}$