4x^2+12x+4=40

Solution for 4x^2+12x+4=40 equation:

4x^2+12x+4=40

We move all terms to the left:

4x^2+12x+4-(40)=0

We add all the numbers together, and all the variables

4x^2+12x-36=0

a = 4; b = 12; c = -36;
Δ = b2-4ac
Δ = 122-4·4·(-36)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{5}}{2*4}=\frac{-12-12\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{5}}{2*4}=\frac{-12+12\sqrt{5}}{8}
$