y^2+68y+32=0

Solution for y^2+68y+32=0 equation:

y^2+68y+32=0

a = 1; b = 68; c = +32;
Δ = b2-4ac
Δ = 682-4·1·32
Δ = 4496
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:

$\sqrt{\Delta}=\sqrt{4496}=\sqrt{16*281}=\sqrt{16}*\sqrt{281}=4\sqrt{281}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-4\sqrt{281}}{2*1}=\frac{-68-4\sqrt{281}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+4\sqrt{281}}{2*1}=\frac{-68+4\sqrt{281}}{2}
$