7y+2y(y-3)=6(y+1)-4

Solution for 7y+2y(y-3)=6(y+1)-4 equation:

7y+2y(y-3)=6(y+1)-4

We move all terms to the left:

7y+2y(y-3)-(6(y+1)-4)=0

We multiply parentheses

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


We calculate terms in parentheses: -(6(y+1)-4), so:

6(y+1)-4

We multiply parentheses

6y+6-4

We add all the numbers together, and all the variables

6y+2

Back to the equation:

-(6y+2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2-5y-2=0

a = 2; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·2·(-2)
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{-(-5)-\sqrt{41}}{2*2}=\frac{5-\sqrt{41}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{41}}{2*2}=\frac{5+\sqrt{41}}{4}
$