x^2+8x=441

Solution for x^2+8x=441 equation:

x^2+8x=441

We move all terms to the left:

x^2+8x-(441)=0

a = 1; b = 8; c = -441;
Δ = b2-4ac
Δ = 82-4·1·(-441)
Δ = 1828
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:

$\sqrt{\Delta}=\sqrt{1828}=\sqrt{4*457}=\sqrt{4}*\sqrt{457}=2\sqrt{457}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{457}}{2*1}=\frac{-8-2\sqrt{457}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{457}}{2*1}=\frac{-8+2\sqrt{457}}{2}
$