6(3-x)+(-20)=10+3(4x2)

Solution for 6(3-x)+(-20)=10+3(4x2) equation:

6(3-x)+(-20)=10+3(4x^2)

We move all terms to the left:

6(3-x)+(-20)-(10+3(4x^2))=0

determiningTheFunctionDomain
-(10+34x^2)+6(3-x)+(-20)=0

We add all the numbers together, and all the variables

-(10+34x^2)+6(-1x+3)+(-20)=0

We add all the numbers together, and all the variables

-(10+34x^2)+6(-1x+3)-20=0

We multiply parentheses

-(10+34x^2)-6x+18-20=0

We get rid of parentheses

-34x^2-6x-10+18-20=0

We add all the numbers together, and all the variables

-34x^2-6x-12=0

a = -34; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·(-34)·(-12)
Δ = -1596
Delta is less than zero, so there is no solution for the equation