(y)^2-5y+6=0

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Solution for (y)^2-5y+6=0 equation:



(y)^2-5y+6=0
a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2} =2 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2} =3 $

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