2x^2+3x=125

Solution for 2x^2+3x=125 equation:

2x^2+3x=125

We move all terms to the left:

2x^2+3x-(125)=0

a = 2; b = 3; c = -125;
Δ = b2-4ac
Δ = 32-4·2·(-125)
Δ = 1009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{1009}}{2*2}=\frac{-3-\sqrt{1009}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1009}}{2*2}=\frac{-3+\sqrt{1009}}{4}
$