(2*k^2-k+1)/2-1=0

Solution for (2*k^2-k+1)/2-1=0 equation:

(2k^2-k+1)/2-1=0

We multiply all the terms by the denominator


(2k^2-k+1)-1*2=0

We add all the numbers together, and all the variables

(2k^2-k+1)-2=0

We get rid of parentheses

2k^2-k+1-2=0

We add all the numbers together, and all the variables

2k^2-1k-1=0

a = 2; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·2·(-1)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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