5w^2+20w-3=0

Solution for 5w^2+20w-3=0 equation:

5w^2+20w-3=0

a = 5; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·5·(-3)
Δ = 460
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:

$\sqrt{\Delta}=\sqrt{460}=\sqrt{4*115}=\sqrt{4}*\sqrt{115}=2\sqrt{115}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{115}}{2*5}=\frac{-20-2\sqrt{115}}{10}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{115}}{2*5}=\frac{-20+2\sqrt{115}}{10}
$