20y^2+11y=420y2+11y=4.

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Solution for 20y^2+11y=420y2+11y=4. equation: 



20y^2+11y=420y^2+11y=4.

We move all terms to the left:

20y^2+11y-(420y^2+11y)=0

We get rid of parentheses

20y^2-420y^2+11y-11y=0

We add all the numbers together, and all the variables

-400y^2=0

a = -400; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-400)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{-800}=0$

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