2r^2+16r-150=-4r

Solution for 2r^2+16r-150=-4r equation:

2r^2+16r-150=-4r

We move all terms to the left:

2r^2+16r-150-(-4r)=0

We get rid of parentheses

2r^2+16r+4r-150=0

We add all the numbers together, and all the variables

2r^2+20r-150=0

a = 2; b = 20; c = -150;
Δ = b2-4ac
Δ = 202-4·2·(-150)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-40}{2*2}=\frac{-60}{4}
=-15
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+40}{2*2}=\frac{20}{4}
=5
$