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

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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

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