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

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

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

We move all terms to the left:

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

Wy multiply elements

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-3=0

a = 2; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*2}=\frac{-4}{4}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*2}=\frac{6}{4}
=1+1/2
$