20=4z^2+Z

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Solution for 20=4z^2+Z equation: 



20=4z^2+

We move all terms to the left:

20-(4z^2+)=0

We get rid of parentheses

-4z^2+20-=0

We add all the numbers together, and all the variables

-4z^2=0

a = -4; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-4)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$z=\frac{-b}{2a}=\frac{0}{-8}=0$

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