2j^2-7j+5=0

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Solution for 2j^2-7j+5=0 equation:



2j^2-7j+5=0
a = 2; b = -7; c = +5;
Δ = b2-4ac
Δ = -72-4·2·5
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*2}=\frac{4}{4} =1 $
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*2}=\frac{10}{4} =2+1/2 $

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