X^2+32x+248=0

Solution for X^2+32x+248=0 equation:

X^2+32X+248=0

a = 1; b = 32; c = +248;
Δ = b2-4ac
Δ = 322-4·1·248
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{2}}{2*1}=\frac{-32-4\sqrt{2}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{2}}{2*1}=\frac{-32+4\sqrt{2}}{2}
$