# How do you solve 3x^2 + 5x + 1 = 0?

Aug 3, 2015

This equation has two real solutions:

${x}_{1} = \frac{- 5 - \sqrt{13}}{6}$

${x}_{2} = \frac{- 5 + \sqrt{13}}{6}$

#### Explanation:

To solve a quadratic equation you have to calculate $\Delta$

$\Delta = {b}^{2} - 4 a c$

$\Delta = {5}^{2} - 4 \cdot 3 \cdot 1 = 25 - 12 = 13$

Calculated value is positive, so the equation has 2 real solutions, which can be calculated using:

${x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{4 a}$