2x^2+3x=40

Solution for 2x^2+3x=40 equation:

2x^2+3x=40

We move all terms to the left:

2x^2+3x-(40)=0

a = 2; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·2·(-40)
Δ = 329
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{329}}{2*2}=\frac{-3-\sqrt{329}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{329}}{2*2}=\frac{-3+\sqrt{329}}{4}
$