4x^2=15x+40

Solution for 4x^2=15x+40 equation:

4x^2=15x+40

We move all terms to the left:

4x^2-(15x+40)=0

We get rid of parentheses

4x^2-15x-40=0

a = 4; b = -15; c = -40;
Δ = b2-4ac
Δ = -152-4·4·(-40)
Δ = 865
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{-(-15)-\sqrt{865}}{2*4}=\frac{15-\sqrt{865}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{865}}{2*4}=\frac{15+\sqrt{865}}{8}
$