12x^2+18x+0=0

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

12x^2+18x+0=0

We add all the numbers together, and all the variables

12x^2+18x=0

a = 12; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·12·0
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{-(18)-18}{2*12}=\frac{-36}{24}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*12}=\frac{0}{24}
=0
$