(6h-4)-(2h+1);h=3

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Solution for (6h-4)-(2h+1);h=3 equation: 



(6h-4)-(2h+1)h=3

We move all terms to the left:

(6h-4)-(2h+1)h-(3)=0

We multiply parentheses

-2h^2+(6h-4)-h-3=0

We get rid of parentheses

-2h^2+6h-h-4-3=0

We add all the numbers together, and all the variables

-2h^2+5h-7=0

a = -2; b = 5; c = -7;
Δ = b2-4ac
Δ = 52-4·(-2)·(-7)
Δ = -31
Delta is less than zero, so there is no solution for the equation

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