28=(7y-4)(y+5)

Solution for 28=(7y-4)(y+5) equation:

28=(7y-4)(y+5)

We move all terms to the left:

28-((7y-4)(y+5))=0

We multiply parentheses ..

-((+7y^2+35y-4y-20))+28=0


We calculate terms in parentheses: -((+7y^2+35y-4y-20)), so:

(+7y^2+35y-4y-20)

We get rid of parentheses

7y^2+35y-4y-20

We add all the numbers together, and all the variables

7y^2+31y-20

Back to the equation:

-(7y^2+31y-20)


We get rid of parentheses

-7y^2-31y+20+28=0

We add all the numbers together, and all the variables

-7y^2-31y+48=0

a = -7; b = -31; c = +48;
Δ = b2-4ac
Δ = -312-4·(-7)·48
Δ = 2305
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{-(-31)-\sqrt{2305}}{2*-7}=\frac{31-\sqrt{2305}}{-14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{2305}}{2*-7}=\frac{31+\sqrt{2305}}{-14}
$