28=y(3y+5)

Solution for 28=y(3y+5) equation:

28=y(3y+5)

We move all terms to the left:

28-(y(3y+5))=0


We calculate terms in parentheses: -(y(3y+5)), so:

y(3y+5)

We multiply parentheses

3y^2+5y

Back to the equation:

-(3y^2+5y)


We get rid of parentheses

-3y^2-5y+28=0

a = -3; b = -5; c = +28;
Δ = b2-4ac
Δ = -52-4·(-3)·28
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*-3}=\frac{-14}{-6}
=2+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*-3}=\frac{24}{-6}
=-4
$