x^2+3x+15=40

Solution for x^2+3x+15=40 equation:

x^2+3x+15=40

We move all terms to the left:

x^2+3x+15-(40)=0

We add all the numbers together, and all the variables

x^2+3x-25=0

a = 1; b = 3; c = -25;
Δ = b2-4ac
Δ = 32-4·1·(-25)
Δ = 109
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{109}}{2*1}=\frac{-3-\sqrt{109}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{109}}{2*1}=\frac{-3+\sqrt{109}}{2}
$