10y-5y(3y+1)=8(y-4)

Solution for 10y-5y(3y+1)=8(y-4) equation:

10y-5y(3y+1)=8(y-4)

We move all terms to the left:

10y-5y(3y+1)-(8(y-4))=0

We multiply parentheses

-15y^2+10y-5y-(8(y-4))=0


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

8(y-4)

We multiply parentheses

8y-32

Back to the equation:

-(8y-32)


We add all the numbers together, and all the variables

-15y^2+5y-(8y-32)=0

We get rid of parentheses

-15y^2+5y-8y+32=0

We add all the numbers together, and all the variables

-15y^2-3y+32=0

a = -15; b = -3; c = +32;
Δ = b2-4ac
Δ = -32-4·(-15)·32
Δ = 1929
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{-(-3)-\sqrt{1929}}{2*-15}=\frac{3-\sqrt{1929}}{-30}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1929}}{2*-15}=\frac{3+\sqrt{1929}}{-30}
$