4k-3=k(2k-1)

Solution for 4k-3=k(2k-1) equation:

4k-3=k(2k-1)

We move all terms to the left:

4k-3-(k(2k-1))=0


We calculate terms in parentheses: -(k(2k-1)), so:

k(2k-1)

We multiply parentheses

2k^2-1k

Back to the equation:

-(2k^2-1k)


We get rid of parentheses

-2k^2+4k+1k-3=0

We add all the numbers together, and all the variables

-2k^2+5k-3=0

a = -2; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·(-2)·(-3)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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