2y^2-20y+28=0

Solution for 2y^2-20y+28=0 equation:

2y^2-20y+28=0

a = 2; b = -20; c = +28;
Δ = b2-4ac
Δ = -202-4·2·28
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{11}}{2*2}=\frac{20-4\sqrt{11}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{11}}{2*2}=\frac{20+4\sqrt{11}}{4}
$