4w^2+15w=550

Solution for 4w^2+15w=550 equation:

4w^2+15w=550

We move all terms to the left:

4w^2+15w-(550)=0

a = 4; b = 15; c = -550;
Δ = b2-4ac
Δ = 152-4·4·(-550)
Δ = 9025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9025}=95$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-95}{2*4}=\frac{-110}{8}
=-13+3/4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+95}{2*4}=\frac{80}{8}
=10
$