j^2-22j-39=0

Solution for j^2-22j-39=0 equation:

j^2-22j-39=0

a = 1; b = -22; c = -39;
Δ = b2-4ac
Δ = -222-4·1·(-39)
Δ = 640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-8\sqrt{10}}{2*1}=\frac{22-8\sqrt{10}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+8\sqrt{10}}{2*1}=\frac{22+8\sqrt{10}}{2}
$