200=-2x^2+40x

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Solution for 200=-2x^2+40x equation: 



200=-2x^2+40x

We move all terms to the left:

200-(-2x^2+40x)=0

We get rid of parentheses

2x^2-40x+200=0

a = 2; b = -40; c = +200;
Δ = b2-4ac
Δ = -402-4·2·200
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{40}{4}=10$

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