13=-2y(y-4)+3y

Solution for 13=-2y(y-4)+3y equation:

13=-2y(y-4)+3y

We move all terms to the left:

13-(-2y(y-4)+3y)=0


We calculate terms in parentheses: -(-2y(y-4)+3y), so:

-2y(y-4)+3y

We add all the numbers together, and all the variables

3y-2y(y-4)

We multiply parentheses

-2y^2+3y+8y

We add all the numbers together, and all the variables

-2y^2+11y

Back to the equation:

-(-2y^2+11y)


We get rid of parentheses

2y^2-11y+13=0

a = 2; b = -11; c = +13;
Δ = b2-4ac
Δ = -112-4·2·13
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{17}}{2*2}=\frac{11-\sqrt{17}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{17}}{2*2}=\frac{11+\sqrt{17}}{4}
$