2y^2+10y=5

Solution for 2y^2+10y=5 equation:

2y^2+10y=5

We move all terms to the left:

2y^2+10y-(5)=0

a = 2; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·2·(-5)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*2}=\frac{-10-2\sqrt{35}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*2}=\frac{-10+2\sqrt{35}}{4}
$