20x=5x^2+20

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Solution for 20x=5x^2+20 equation: 



20x=5x^2+20

We move all terms to the left:

20x-(5x^2+20)=0

We get rid of parentheses

-5x^2+20x-20=0

a = -5; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·(-5)·(-20)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-20}{-10}=+2$

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