2x^2+4x+4=41

Solution for 2x^2+4x+4=41 equation:

2x^2+4x+4=41

We move all terms to the left:

2x^2+4x+4-(41)=0

We add all the numbers together, and all the variables

2x^2+4x-37=0

a = 2; b = 4; c = -37;
Δ = b2-4ac
Δ = 42-4·2·(-37)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{78}}{2*2}=\frac{-4-2\sqrt{78}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{78}}{2*2}=\frac{-4+2\sqrt{78}}{4}
$