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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+3y^2+15y+2y+10)-(-2(3y+2))=0


We calculate terms in parentheses: -(-2(3y+2)), so:

-2(3y+2)

We multiply parentheses

-6y-4

Back to the equation:

-(-6y-4)


We get rid of parentheses

3y^2+15y+2y+6y+10+4=0

We add all the numbers together, and all the variables

3y^2+23y+14=0

a = 3; b = 23; c = +14;
Δ = b2-4ac
Δ = 232-4·3·14
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-19}{2*3}=\frac{-42}{6}
=-7
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+19}{2*3}=\frac{-4}{6}
=-2/3
$