2y+5y(y-6)=8y-3y(y-10)

Solution for 2y+5y(y-6)=8y-3y(y-10) equation:

2y+5y(y-6)=8y-3y(y-10)

We move all terms to the left:

2y+5y(y-6)-(8y-3y(y-10))=0

We multiply parentheses

5y^2+2y-30y-(8y-3y(y-10))=0


We calculate terms in parentheses: -(8y-3y(y-10)), so:

8y-3y(y-10)

We multiply parentheses

-3y^2+8y+30y

We add all the numbers together, and all the variables

-3y^2+38y

Back to the equation:

-(-3y^2+38y)


We add all the numbers together, and all the variables

5y^2-(-3y^2+38y)-28y=0

We get rid of parentheses

5y^2+3y^2-38y-28y=0

We add all the numbers together, and all the variables

8y^2-66y=0

a = 8; b = -66; c = 0;
Δ = b2-4ac
Δ = -662-4·8·0
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-66)-66}{2*8}=\frac{0}{16}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-66)+66}{2*8}=\frac{132}{16}
=8+1/4
$