y^2+18y+12=0

Solution for y^2+18y+12=0 equation:

y^2+18y+12=0

a = 1; b = 18; c = +12;
Δ = b2-4ac
Δ = 182-4·1·12
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{69}}{2*1}=\frac{-18-2\sqrt{69}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{69}}{2*1}=\frac{-18+2\sqrt{69}}{2}
$