68=(6x^2)+(32x+40)

Solution for 68=(6x^2)+(32x+40) equation:

68=(6x^2)+(32x+40)

We move all terms to the left:

68-((6x^2)+(32x+40))=0


We calculate terms in parentheses: -(6x^2+(32x+40)), so:

6x^2+(32x+40)

We get rid of parentheses

6x^2+32x+40

Back to the equation:

-(6x^2+32x+40)


We get rid of parentheses

-6x^2-32x-40+68=0

We add all the numbers together, and all the variables

-6x^2-32x+28=0

a = -6; b = -32; c = +28;
Δ = b2-4ac
Δ = -322-4·(-6)·28
Δ = 1696
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{106}}{2*-6}=\frac{32-4\sqrt{106}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{106}}{2*-6}=\frac{32+4\sqrt{106}}{-12}
$