x^2+21x=180

Solution for x^2+21x=180 equation:

x^2+21x=180

We move all terms to the left:

x^2+21x-(180)=0

a = 1; b = 21; c = -180;
Δ = b2-4ac
Δ = 212-4·1·(-180)
Δ = 1161
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:

$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{129}}{2*1}=\frac{-21-3\sqrt{129}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{129}}{2*1}=\frac{-21+3\sqrt{129}}{2}
$