(5y+4)(y+4)=-3(5y+4)

Solution for (5y+4)(y+4)=-3(5y+4) equation:

(5y+4)(y+4)=-3(5y+4)

We move all terms to the left:

(5y+4)(y+4)-(-3(5y+4))=0

We multiply parentheses ..

(+5y^2+20y+4y+16)-(-3(5y+4))=0


We calculate terms in parentheses: -(-3(5y+4)), so:

-3(5y+4)

We multiply parentheses

-15y-12

Back to the equation:

-(-15y-12)


We get rid of parentheses

5y^2+20y+4y+15y+16+12=0

We add all the numbers together, and all the variables

5y^2+39y+28=0

a = 5; b = 39; c = +28;
Δ = b2-4ac
Δ = 392-4·5·28
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-31}{2*5}=\frac{-70}{10}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+31}{2*5}=\frac{-8}{10}
=-4/5
$