q^2-18q+97=16

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Solution for q^2-18q+97=16 equation: 



q^2-18q+97=16

We move all terms to the left:

q^2-18q+97-(16)=0

We add all the numbers together, and all the variables

q^2-18q+81=0

a = 1; b = -18; c = +81;
Δ = b2-4ac
Δ = -182-4·1·81
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$q=\frac{-b}{2a}=\frac{18}{2}=9$

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