X^2+51x+540=0

Solution for X^2+51x+540=0 equation:

X^2+51X+540=0

a = 1; b = 51; c = +540;
Δ = b2-4ac
Δ = 512-4·1·540
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-21}{2*1}=\frac{-72}{2}
=-36
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+21}{2*1}=\frac{-30}{2}
=-15
$