x^2+4x+21=180

Solution for x^2+4x+21=180 equation:

x^2+4x+21=180

We move all terms to the left:

x^2+4x+21-(180)=0

We add all the numbers together, and all the variables

x^2+4x-159=0

a = 1; b = 4; c = -159;
Δ = b2-4ac
Δ = 42-4·1·(-159)
Δ = 652
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:

$\sqrt{\Delta}=\sqrt{652}=\sqrt{4*163}=\sqrt{4}*\sqrt{163}=2\sqrt{163}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{163}}{2*1}=\frac{-4-2\sqrt{163}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{163}}{2*1}=\frac{-4+2\sqrt{163}}{2}
$