x^2+10x+21=396

Solution for x^2+10x+21=396 equation:

x^2+10x+21=396

We move all terms to the left:

x^2+10x+21-(396)=0

We add all the numbers together, and all the variables

x^2+10x-375=0

a = 1; b = 10; c = -375;
Δ = b2-4ac
Δ = 102-4·1·(-375)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-40}{2*1}=\frac{-50}{2}
=-25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+40}{2*1}=\frac{30}{2}
=15
$