242=4x^2+12x

Solution for 242=4x^2+12x equation:

242=4x^2+12x

We move all terms to the left:

242-(4x^2+12x)=0

We get rid of parentheses

-4x^2-12x+242=0

a = -4; b = -12; c = +242;
Δ = b2-4ac
Δ = -122-4·(-4)·242
Δ = 4016
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4016}=\sqrt{16*251}=\sqrt{16}*\sqrt{251}=4\sqrt{251}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{251}}{2*-4}=\frac{12-4\sqrt{251}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{251}}{2*-4}=\frac{12+4\sqrt{251}}{-8}
$