10x^2+20x=150

Solution for 10x^2+20x=150 equation:

10x^2+20x=150

We move all terms to the left:

10x^2+20x-(150)=0

a = 10; b = 20; c = -150;
Δ = b2-4ac
Δ = 202-4·10·(-150)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-80}{2*10}=\frac{-100}{20}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+80}{2*10}=\frac{60}{20}
=3
$