0.5x^2+20x=2250

Solution for 0.5x^2+20x=2250 equation:

0.5x^2+20x=2250

We move all terms to the left:

0.5x^2+20x-(2250)=0

a = 0.5; b = 20; c = -2250;
Δ = b2-4ac
Δ = 202-4·0.5·(-2250)
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-70}{2*0.5}=\frac{-90}{1}
=-90
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+70}{2*0.5}=\frac{50}{1}
=50
$