2y^2+20y=18

Solution for 2y^2+20y=18 equation:

2y^2+20y=18

We move all terms to the left:

2y^2+20y-(18)=0

a = 2; b = 20; c = -18;
Δ = b2-4ac
Δ = 202-4·2·(-18)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{34}}{2*2}=\frac{-20-4\sqrt{34}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{34}}{2*2}=\frac{-20+4\sqrt{34}}{4}
$