12(3+y)=5(y2+8)

Solution for 12(3+y)=5(y2+8) equation:

12(3+y)=5(y2+8)

We move all terms to the left:

12(3+y)-(5(y2+8))=0

We add all the numbers together, and all the variables

-(5(+y^2+8))+12(y+3)=0

We multiply parentheses

-(5(+y^2+8))+12y+36=0


We calculate terms in parentheses: -(5(+y^2+8)), so:

5(+y^2+8)

We multiply parentheses

5y^2+40

Back to the equation:

-(5y^2+40)


We add all the numbers together, and all the variables

12y-(5y^2+40)+36=0

We get rid of parentheses

-5y^2+12y-40+36=0

We add all the numbers together, and all the variables

-5y^2+12y-4=0

a = -5; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·(-5)·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*-5}=\frac{-20}{-10}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*-5}=\frac{-4}{-10}
=2/5
$