X^2+22x-5040=0

Solution for X^2+22x-5040=0 equation:

X^2+22X-5040=0

a = 1; b = 22; c = -5040;
Δ = b2-4ac
Δ = 222-4·1·(-5040)
Δ = 20644
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:

$\sqrt{\Delta}=\sqrt{20644}=\sqrt{4*5161}=\sqrt{4}*\sqrt{5161}=2\sqrt{5161}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{5161}}{2*1}=\frac{-22-2\sqrt{5161}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{5161}}{2*1}=\frac{-22+2\sqrt{5161}}{2}
$