y^2+18y-3=0

Solution for y^2+18y-3=0 equation:

y^2+18y-3=0

a = 1; b = 18; c = -3;
Δ = b2-4ac
Δ = 182-4·1·(-3)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{21}}{2*1}=\frac{-18-4\sqrt{21}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{21}}{2*1}=\frac{-18+4\sqrt{21}}{2}
$