3x^2+12x+x=0

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

3x^2+12x+x=0

We add all the numbers together, and all the variables

3x^2+13x=0

a = 3; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·3·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*3}=\frac{-26}{6}
=-4+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*3}=\frac{0}{6}
=0
$