x^2+3x-40=68

Solution for x^2+3x-40=68 equation:

x^2+3x-40=68

We move all terms to the left:

x^2+3x-40-(68)=0

We add all the numbers together, and all the variables

x^2+3x-108=0

a = 1; b = 3; c = -108;
Δ = b2-4ac
Δ = 32-4·1·(-108)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*1}=\frac{18}{2}
=9
$