(2y+3)(3y-4)=-2y(6y+13y)-16

Solution for (2y+3)(3y-4)=-2y(6y+13y)-16 equation:

(2y+3)(3y-4)=-2y(6y+13y)-16

We move all terms to the left:

(2y+3)(3y-4)-(-2y(6y+13y)-16)=0

We add all the numbers together, and all the variables

(2y+3)(3y-4)-(-2y(+19y)-16)=0

We multiply parentheses ..

(+6y^2-8y+9y-12)-(-2y(+19y)-16)=0


We calculate terms in parentheses: -(-2y(+19y)-16), so:

-2y(+19y)-16

We multiply parentheses

-38y^2-16

Back to the equation:

-(-38y^2-16)


We get rid of parentheses

6y^2+38y^2-8y+9y-12+16=0

We add all the numbers together, and all the variables

44y^2+y+4=0

a = 44; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·44·4
Δ = -703
Delta is less than zero, so there is no solution for the equation