20*x^2-5*x=0

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Solution for 20*x^2-5*x=0 equation: 



20x^2-5x=0

a = 20; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·20·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*20}=\frac{0}{40}
=0
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*20}=\frac{10}{40}
=1/4
$

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