X^2+52x+504=0

Solution for X^2+52x+504=0 equation:

X^2+52X+504=0

a = 1; b = 52; c = +504;
Δ = b2-4ac
Δ = 522-4·1·504
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{43}}{2*1}=\frac{-52-4\sqrt{43}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{43}}{2*1}=\frac{-52+4\sqrt{43}}{2}
$