3x^2+12x=400

Solution for 3x^2+12x=400 equation:

3x^2+12x=400

We move all terms to the left:

3x^2+12x-(400)=0

a = 3; b = 12; c = -400;
Δ = b2-4ac
Δ = 122-4·3·(-400)
Δ = 4944
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:

$\sqrt{\Delta}=\sqrt{4944}=\sqrt{16*309}=\sqrt{16}*\sqrt{309}=4\sqrt{309}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{309}}{2*3}=\frac{-12-4\sqrt{309}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{309}}{2*3}=\frac{-12+4\sqrt{309}}{6}
$