2w^2+20w-36=0

Solution for 2w^2+20w-36=0 equation:

2w^2+20w-36=0

a = 2; b = 20; c = -36;
Δ = b2-4ac
Δ = 202-4·2·(-36)
Δ = 688
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{43}}{2*2}=\frac{-20-4\sqrt{43}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{43}}{2*2}=\frac{-20+4\sqrt{43}}{4}
$