(4q+5q)(4q+5q)=0

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Solution for (4q+5q)(4q+5q)=0 equation: 



(4q+5q)(4q+5q)=0

We add all the numbers together, and all the variables

(+9q)(+9q)=0

We multiply parentheses ..

(+81q^2)=0

We get rid of parentheses

81q^2=0

a = 81; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·81·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$q=\frac{-b}{2a}=\frac{0}{162}=0$

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