x^2+21x+32=20

Solution for x^2+21x+32=20 equation:

x^2+21x+32=20

We move all terms to the left:

x^2+21x+32-(20)=0

We add all the numbers together, and all the variables

x^2+21x+12=0

a = 1; b = 21; c = +12;
Δ = b2-4ac
Δ = 212-4·1·12
Δ = 393
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{393}}{2*1}=\frac{-21-\sqrt{393}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{393}}{2*1}=\frac{-21+\sqrt{393}}{2}
$