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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+4y^2+8y+5y+10)-(-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+8y+5y+20y+10+25=0

We add all the numbers together, and all the variables

4y^2+33y+35=0

a = 4; b = 33; c = +35;
Δ = b2-4ac
Δ = 332-4·4·35
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-23}{2*4}=\frac{-56}{8}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+23}{2*4}=\frac{-10}{8}
=-1+1/4
$