# How do you use the quadratic formula to solve 5s^2+6x+3=0?

Jan 30, 2018

x = 0.2*(-3 + sqrt6color(red) i), (0.2 * (-3 - sqrt6color(red) i)

$x = 0.2 \cdot \left(- 3 \pm 2.4495 \textcolor{red}{i}\right)$

#### Explanation:

Given Equation : $5 {x}^{2} + 6 x + 3 = 0$

Formula for finding the routes : $\frac{- b \pm \sqrt{{b}^{2} - \left(4 a c\right)}}{2 a}$

where a, b c are the coefficients of ${x}^{2} , x \mathmr{and} c o n s t$ terms respy.

$x = \frac{- 6 \pm \sqrt{{6}^{2} - \left(4 \cdot 5 \cdot 3\right)}}{2 \cdot 5}$

$x = \frac{- 6 \pm \sqrt{- 24}}{10}$

$x = - 0.6 \pm 0.2 \sqrt{6} i$

$x = - 0.6 + 0.2 \sqrt{6} \textcolor{red}{i} , - 0.6 - 0.2 \sqrt{6} \textcolor{red}{i}$