4q^+3q=3q^2-4q+18

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Solution for 4q^+3q=3q^2-4q+18 equation: 



4q^+3q=3q^2-4q+18

We move all terms to the left:

4q^+3q-(3q^2-4q+18)=0

We add all the numbers together, and all the variables

7q-(3q^2-4q+18)=0

We get rid of parentheses

-3q^2+7q+4q-18=0

We add all the numbers together, and all the variables

-3q^2+11q-18=0

a = -3; b = 11; c = -18;
Δ = b2-4ac
Δ = 112-4·(-3)·(-18)
Δ = -95
Delta is less than zero, so there is no solution for the equation

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