r(2)+25r-150=0

Solution for r(2)+25r-150=0 equation:

r(2)+25r-150=0

We add all the numbers together, and all the variables

r^2+25r-150=0

a = 1; b = 25; c = -150;
Δ = b2-4ac
Δ = 252-4·1·(-150)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-35}{2*1}=\frac{-60}{2}
=-30
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+35}{2*1}=\frac{10}{2}
=5
$