7y+2y(y-2)=6(y+1)-2

Solution for 7y+2y(y-2)=6(y+1)-2 equation:

7y+2y(y-2)=6(y+1)-2

We move all terms to the left:

7y+2y(y-2)-(6(y+1)-2)=0

We multiply parentheses

2y^2+7y-4y-(6(y+1)-2)=0


We calculate terms in parentheses: -(6(y+1)-2), so:

6(y+1)-2

We multiply parentheses

6y+6-2

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

2y^2+3y-(6y+4)=0

We get rid of parentheses

2y^2+3y-6y-4=0

We add all the numbers together, and all the variables

2y^2-3y-4=0

a = 2; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·2·(-4)
Δ = 41
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{41}}{2*2}=\frac{3-\sqrt{41}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*2}=\frac{3+\sqrt{41}}{4}
$