2(5c+2)(-2c)=3(2c+3)+7

Solution for 2(5c+2)(-2c)=3(2c+3)+7 equation:

2(5c+2)(-2c)=3(2c+3)+7

We move all terms to the left:

2(5c+2)(-2c)-(3(2c+3)+7)=0

We multiply parentheses ..

2(-10c^2-4c)-(3(2c+3)+7)=0


We calculate terms in parentheses: -(3(2c+3)+7), so:

3(2c+3)+7

We multiply parentheses

6c+9+7

We add all the numbers together, and all the variables

6c+16

Back to the equation:

-(6c+16)


We multiply parentheses

-20c^2-8c-(6c+16)=0

We get rid of parentheses

-20c^2-8c-6c-16=0

We add all the numbers together, and all the variables

-20c^2-14c-16=0

a = -20; b = -14; c = -16;
Δ = b2-4ac
Δ = -142-4·(-20)·(-16)
Δ = -1084
Delta is less than zero, so there is no solution for the equation