b^2+20b-40=0

Solution for b^2+20b-40=0 equation:

b^2+20b-40=0

a = 1; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·1·(-40)
Δ = 560
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{35}}{2*1}=\frac{-20-4\sqrt{35}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{35}}{2*1}=\frac{-20+4\sqrt{35}}{2}
$