3x^2+36x+49=8

Solution for 3x^2+36x+49=8 equation:

3x^2+36x+49=8

We move all terms to the left:

3x^2+36x+49-(8)=0

We add all the numbers together, and all the variables

3x^2+36x+41=0

a = 3; b = 36; c = +41;
Δ = b2-4ac
Δ = 362-4·3·41
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{201}}{2*3}=\frac{-36-2\sqrt{201}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{201}}{2*3}=\frac{-36+2\sqrt{201}}{6}
$