8x^2+32x+29=0

Solution for 8x^2+32x+29=0 equation:

8x^2+32x+29=0

a = 8; b = 32; c = +29;
Δ = b2-4ac
Δ = 322-4·8·29
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{6}}{2*8}=\frac{-32-4\sqrt{6}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{6}}{2*8}=\frac{-32+4\sqrt{6}}{16}
$