-1+3u-2u^2=0

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Solution for -1+3u-2u^2=0 equation:



-1+3u-2u^2=0
a = -2; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·(-2)·(-1)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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

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