0=8(x^2+4x)+17

Solution for 0=8(x^2+4x)+17 equation:

0=8(x^2+4x)+17

We move all terms to the left:

0-(8(x^2+4x)+17)=0

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(8(x^2+4x)+17), so:

8(x^2+4x)+17

We multiply parentheses

8x^2+32x+17

Back to the equation:

-(8x^2+32x+17)


We get rid of parentheses

-8x^2-32x-17=0

a = -8; b = -32; c = -17;
Δ = b2-4ac
Δ = -322-4·(-8)·(-17)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{30}}{2*-8}=\frac{32-4\sqrt{30}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{30}}{2*-8}=\frac{32+4\sqrt{30}}{-16}
$