q^2-9q+20=0

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Solution for q^2-9q+20=0 equation: 



q^2-9q+20=0

a = 1; b = -9; c = +20;
Δ = b2-4ac
Δ = -92-4·1·20
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*1}=\frac{8}{2}
=4
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*1}=\frac{10}{2}
=5
$

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