-15+3x^2+12x=0

Solution for -15+3x^2+12x=0 equation:

-15+3x^2+12x=0

a = 3; b = 12; c = -15;
Δ = b2-4ac
Δ = 122-4·3·(-15)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18}{2*3}=\frac{-30}{6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18}{2*3}=\frac{6}{6}
=1
$