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 = 4761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4761}=69$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-69}{2*1}=\frac{-120}{2}
=-60
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+69}{2*1}=\frac{18}{2}
=9
$