2y^2+18y-10=0

Solution for 2y^2+18y-10=0 equation:

2y^2+18y-10=0

a = 2; b = 18; c = -10;
Δ = b2-4ac
Δ = 182-4·2·(-10)
Δ = 404
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:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{101}}{2*2}=\frac{-18-2\sqrt{101}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{101}}{2*2}=\frac{-18+2\sqrt{101}}{4}
$