2x^2+4x+4=17

Solution for 2x^2+4x+4=17 equation:

2x^2+4x+4=17

We move all terms to the left:

2x^2+4x+4-(17)=0

We add all the numbers together, and all the variables

2x^2+4x-13=0

a = 2; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·2·(-13)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{30}}{2*2}=\frac{-4-2\sqrt{30}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{30}}{2*2}=\frac{-4+2\sqrt{30}}{4}
$