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

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

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

We move all terms to the left:

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

We multiply parentheses

-4y^2-16y-(3(y+4)+7)=0


We calculate terms in parentheses: -(3(y+4)+7), so:

3(y+4)+7

We multiply parentheses

3y+12+7

We add all the numbers together, and all the variables

3y+19

Back to the equation:

-(3y+19)


We get rid of parentheses

-4y^2-16y-3y-19=0

We add all the numbers together, and all the variables

-4y^2-19y-19=0

a = -4; b = -19; c = -19;
Δ = b2-4ac
Δ = -192-4·(-4)·(-19)
Δ = 57
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{-(-19)-\sqrt{57}}{2*-4}=\frac{19-\sqrt{57}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{57}}{2*-4}=\frac{19+\sqrt{57}}{-8}
$