16=8y+4y(y-5)

Solution for 16=8y+4y(y-5) equation:

16=8y+4y(y-5)

We move all terms to the left:

16-(8y+4y(y-5))=0


We calculate terms in parentheses: -(8y+4y(y-5)), so:

8y+4y(y-5)

We multiply parentheses

4y^2+8y-20y

We add all the numbers together, and all the variables

4y^2-12y

Back to the equation:

-(4y^2-12y)


We get rid of parentheses

-4y^2+12y+16=0

a = -4; b = 12; c = +16;
Δ = b2-4ac
Δ = 122-4·(-4)·16
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{-(12)-20}{2*-4}=\frac{-32}{-8}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*-4}=\frac{8}{-8}
=-1
$