4f^2=+12f

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

4f^2=+12f

We move all terms to the left:

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

We get rid of parentheses

4f^2-12f=0

a = 4; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·4·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*4}=\frac{0}{8}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*4}=\frac{24}{8}
=3
$