3y-4y(y+1)=-(7y+4)+y

Solution for 3y-4y(y+1)=-(7y+4)+y equation:

3y-4y(y+1)=-(7y+4)+y

We move all terms to the left:

3y-4y(y+1)-(-(7y+4)+y)=0

We multiply parentheses

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


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

-(7y+4)+y

We add all the numbers together, and all the variables

y-(7y+4)

We get rid of parentheses

y-7y-4

We add all the numbers together, and all the variables

-6y-4

Back to the equation:

-(-6y-4)


We add all the numbers together, and all the variables

-4y^2-1y-(-6y-4)=0

We get rid of parentheses

-4y^2-1y+6y+4=0

We add all the numbers together, and all the variables

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

a = -4; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-4)·4
Δ = 89
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{-(5)-\sqrt{89}}{2*-4}=\frac{-5-\sqrt{89}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{89}}{2*-4}=\frac{-5+\sqrt{89}}{-8}
$