5(y+1)+2=4y(y+2)-6

Solution for 5(y+1)+2=4y(y+2)-6 equation:

5(y+1)+2=4y(y+2)-6

We move all terms to the left:

5(y+1)+2-(4y(y+2)-6)=0

We multiply parentheses

5y-(4y(y+2)-6)+5+2=0


We calculate terms in parentheses: -(4y(y+2)-6), so:

4y(y+2)-6

We multiply parentheses

4y^2+8y-6

Back to the equation:

-(4y^2+8y-6)


We add all the numbers together, and all the variables

5y-(4y^2+8y-6)+7=0

We get rid of parentheses

-4y^2+5y-8y+6+7=0

We add all the numbers together, and all the variables

-4y^2-3y+13=0

a = -4; b = -3; c = +13;
Δ = b2-4ac
Δ = -32-4·(-4)·13
Δ = 217
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{217}}{2*-4}=\frac{3-\sqrt{217}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{217}}{2*-4}=\frac{3+\sqrt{217}}{-8}
$