(4y+5)(y+2)=-3(4y+5)

Solution for (4y+5)(y+2)=-3(4y+5) equation:

(4y+5)(y+2)=-3(4y+5)

We move all terms to the left:

(4y+5)(y+2)-(-3(4y+5))=0

We multiply parentheses ..

(+4y^2+8y+5y+10)-(-3(4y+5))=0


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

-3(4y+5)

We multiply parentheses

-12y-15

Back to the equation:

-(-12y-15)


We get rid of parentheses

4y^2+8y+5y+12y+10+15=0

We add all the numbers together, and all the variables

4y^2+25y+25=0

a = 4; b = 25; c = +25;
Δ = b2-4ac
Δ = 252-4·4·25
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15}{2*4}=\frac{-40}{8}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15}{2*4}=\frac{-10}{8}
=-1+1/4
$