j^2-11j+10=0

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Solution for j^2-11j+10=0 equation:



j^2-11j+10=0
a = 1; b = -11; c = +10;
Δ = b2-4ac
Δ = -112-4·1·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-9}{2*1}=\frac{2}{2} =1 $
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+9}{2*1}=\frac{20}{2} =10 $

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