10y2+()=20y

Solution for 10y2+()=20y equation:

10y^2+()=20y

We move all terms to the left:

10y^2+()-(20y)=0

We add all the numbers together, and all the variables

10y^2-20y=0

a = 10; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·10·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*10}=\frac{0}{20}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*10}=\frac{40}{20}
=2
$