Q=100(10(2)^2+(1x2))

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Solution for Q=100(10(2)^2+(1x2)) equation: 



=100(10(2)^2+(1Q^2))

We move all terms to the left:

-(100(10(2)^2+(1Q^2)))=0



We calculate terms in parentheses: -(100(102^2+1Q^2)), so:

100(102^2+1Q^2)

We multiply parentheses

100Q^2+1040400

Back to the equation:

-(100Q^2+1040400)



We get rid of parentheses

-100Q^2-1040400=0

a = -100; b = 0; c = -1040400;
Δ = b2-4ac
Δ = 02-4·(-100)·(-1040400)
Δ = -416160000
Delta is less than zero, so there is no solution for the equation

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