2y+3y(y-4)=7y-4(y-10)

Solution for 2y+3y(y-4)=7y-4(y-10) equation:

2y+3y(y-4)=7y-4(y-10)

We move all terms to the left:

2y+3y(y-4)-(7y-4(y-10))=0

We multiply parentheses

3y^2+2y-12y-(7y-4(y-10))=0


We calculate terms in parentheses: -(7y-4(y-10)), so:

7y-4(y-10)

We multiply parentheses

7y-4y+40

We add all the numbers together, and all the variables

3y+40

Back to the equation:

-(3y+40)


We add all the numbers together, and all the variables

3y^2-10y-(3y+40)=0

We get rid of parentheses

3y^2-10y-3y-40=0

We add all the numbers together, and all the variables

3y^2-13y-40=0

a = 3; b = -13; c = -40;
Δ = b2-4ac
Δ = -132-4·3·(-40)
Δ = 649
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{-(-13)-\sqrt{649}}{2*3}=\frac{13-\sqrt{649}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{649}}{2*3}=\frac{13+\sqrt{649}}{6}
$