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

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

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

We multiply 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}
$