8y^2+28y+8=0

Solution for 8y^2+28y+8=0 equation:

8y^2+28y+8=0

a = 8; b = 28; c = +8;
Δ = b2-4ac
Δ = 282-4·8·8
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{33}}{2*8}=\frac{-28-4\sqrt{33}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{33}}{2*8}=\frac{-28+4\sqrt{33}}{16}
$