9(y+4)=5(y-4)y=

Solution for 9(y+4)=5(y-4)y= equation:

9(y+4)=5(y-4)y=

We move all terms to the left:

9(y+4)-(5(y-4)y)=0

We multiply parentheses

9y-(5(y-4)y)+36=0


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

5(y-4)y

We multiply parentheses

5y^2-20y

Back to the equation:

-(5y^2-20y)


We get rid of parentheses

-5y^2+9y+20y+36=0

We add all the numbers together, and all the variables

-5y^2+29y+36=0

a = -5; b = 29; c = +36;
Δ = b2-4ac
Δ = 292-4·(-5)·36
Δ = 1561
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{-(29)-\sqrt{1561}}{2*-5}=\frac{-29-\sqrt{1561}}{-10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{1561}}{2*-5}=\frac{-29+\sqrt{1561}}{-10}
$