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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+5y^2+15y+4y+12)-(-2(5y+4))=0


We calculate terms in parentheses: -(-2(5y+4)), so:

-2(5y+4)

We multiply parentheses

-10y-8

Back to the equation:

-(-10y-8)


We get rid of parentheses

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

We add all the numbers together, and all the variables

5y^2+29y+20=0

a = 5; b = 29; c = +20;
Δ = b2-4ac
Δ = 292-4·5·20
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*5}=\frac{-50}{10}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*5}=\frac{-8}{10}
=-4/5
$