6x^2+2x+4=40

Solution for 6x^2+2x+4=40 equation:

6x^2+2x+4=40

We move all terms to the left:

6x^2+2x+4-(40)=0

We add all the numbers together, and all the variables

6x^2+2x-36=0

a = 6; b = 2; c = -36;
Δ = b2-4ac
Δ = 22-4·6·(-36)
Δ = 868
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:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{217}}{2*6}=\frac{-2-2\sqrt{217}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{217}}{2*6}=\frac{-2+2\sqrt{217}}{12}
$