(4y+5)(y+7)=-5(4y+5)

Solution for (4y+5)(y+7)=-5(4y+5) equation:

(4y+5)(y+7)=-5(4y+5)

We move all terms to the left:

(4y+5)(y+7)-(-5(4y+5))=0

We multiply parentheses ..

(+4y^2+28y+5y+35)-(-5(4y+5))=0


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

-5(4y+5)

We multiply parentheses

-20y-25

Back to the equation:

-(-20y-25)


We get rid of parentheses

4y^2+28y+5y+20y+35+25=0

We add all the numbers together, and all the variables

4y^2+53y+60=0

a = 4; b = 53; c = +60;
Δ = b2-4ac
Δ = 532-4·4·60
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-43}{2*4}=\frac{-96}{8}
=-12
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+43}{2*4}=\frac{-10}{8}
=-1+1/4
$