(2k-10)(k-5)=0

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Solution for (2k-10)(k-5)=0 equation: 



(2k-10)(k-5)=0

We multiply parentheses ..

(+2k^2-10k-10k+50)=0

We get rid of parentheses

2k^2-10k-10k+50=0

We add all the numbers together, and all the variables

2k^2-20k+50=0

a = 2; b = -20; c = +50;
Δ = b2-4ac
Δ = -202-4·2·50
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$k=\frac{-b}{2a}=\frac{20}{4}=5$

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