4w^2+100w-600=0

Solution for 4w^2+100w-600=0 equation:

4w^2+100w-600=0

a = 4; b = 100; c = -600;
Δ = b2-4ac
Δ = 1002-4·4·(-600)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-140}{2*4}=\frac{-240}{8}
=-30
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+140}{2*4}=\frac{40}{8}
=5
$