((x^2)+4x)=300

Solution for ((x^2)+4x)=300 equation:

((x^2)+4x)=300

We move all terms to the left:

((x^2)+4x)-(300)=0

We get rid of parentheses

x^2+4x-300=0

a = 1; b = 4; c = -300;
Δ = b2-4ac
Δ = 42-4·1·(-300)
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{19}}{2*1}=\frac{-4-8\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{19}}{2*1}=\frac{-4+8\sqrt{19}}{2}
$