# How do you solve 3x^2 + 2x - 5 = 0?

May 7, 2016

$\left(3 x + 5\right) = 0$

$x = - \left(\frac{5}{3}\right)$

$x - 1 = 0$
$x = 1$

Hence, solution is $x = - \left(\frac{5}{3}\right) \mathmr{and} x = 1$

#### Explanation:

The question gives you an equation $3 {x}^{2} + 2 x - 5 = 0$.

Hence to solve the equation, factorise.

$3 {x}^{2} + 2 x - 5 = 0$
$\left(3 x + 5\right) \left(x - 1\right) = 0$

The 2 factors of the equation are $\left(3 x + 2\right) \mathmr{and} \left(x - 1\right) .$

With $\left(3 x + 5\right) = 0$ and $\left(x - 1\right) = 0$
Solve for $x$ to find the solutions.