-r^2+12r-36=0

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Solution for -r^2+12r-36=0 equation: 



-r^2+12r-36=0

We add all the numbers together, and all the variables

-1r^2+12r-36=0

a = -1; b = 12; c = -36;
Δ = b2-4ac
Δ = 122-4·(-1)·(-36)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$r=\frac{-b}{2a}=\frac{-12}{-2}=+6$

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