X(2x+4)=17

Solution for X(2x+4)=17 equation:

X(2X+4)=17

We move all terms to the left:

X(2X+4)-(17)=0

We multiply parentheses

2X^2+4X-17=0

a = 2; b = 4; c = -17;
Δ = b2-4ac
Δ = 42-4·2·(-17)
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{38}}{2*2}=\frac{-4-2\sqrt{38}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{38}}{2*2}=\frac{-4+2\sqrt{38}}{4}
$