4y^2=10y-5

Solution for 4y^2=10y-5 equation:

4y^2=10y-5

We move all terms to the left:

4y^2-(10y-5)=0

We get rid of parentheses

4y^2-10y+5=0

a = 4; b = -10; c = +5;
Δ = b2-4ac
Δ = -102-4·4·5
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{5}}{2*4}=\frac{10-2\sqrt{5}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*4}=\frac{10+2\sqrt{5}}{8}
$