12x^2=4x+

Solution for 12x^2=4x+ equation:

12x^2=4x+

We move all terms to the left:

12x^2-(4x+)=0

We add all the numbers together, and all the variables

12x^2-(+4x)=0

We get rid of parentheses

12x^2-4x=0

a = 12; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·12·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*12}=\frac{0}{24}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*12}=\frac{8}{24}
=1/3
$