3x^2+18x-13=0

Solution for 3x^2+18x-13=0 equation:

3x^2+18x-13=0

a = 3; b = 18; c = -13;
Δ = b2-4ac
Δ = 182-4·3·(-13)
Δ = 480
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{30}}{2*3}=\frac{-18-4\sqrt{30}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{30}}{2*3}=\frac{-18+4\sqrt{30}}{6}
$