12x^2+18x-390=0

Solution for 12x^2+18x-390=0 equation:

12x^2+18x-390=0

a = 12; b = 18; c = -390;
Δ = b2-4ac
Δ = 182-4·12·(-390)
Δ = 19044
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19044}=138$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-138}{2*12}=\frac{-156}{24}
=-6+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+138}{2*12}=\frac{120}{24}
=5
$