14a(1+5a)=4(3a-11)-20

Solution for 14a(1+5a)=4(3a-11)-20 equation:

14a(1+5a)=4(3a-11)-20

We move all terms to the left:

14a(1+5a)-(4(3a-11)-20)=0

We add all the numbers together, and all the variables

14a(5a+1)-(4(3a-11)-20)=0

We multiply parentheses

70a^2+14a-(4(3a-11)-20)=0


We calculate terms in parentheses: -(4(3a-11)-20), so:

4(3a-11)-20

We multiply parentheses

12a-44-20

We add all the numbers together, and all the variables

12a-64

Back to the equation:

-(12a-64)


We get rid of parentheses

70a^2+14a-12a+64=0

We add all the numbers together, and all the variables

70a^2+2a+64=0

a = 70; b = 2; c = +64;
Δ = b2-4ac
Δ = 22-4·70·64
Δ = -17916
Delta is less than zero, so there is no solution for the equation