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

Solution for K^3-4k^2+3k+2=0 equation:

^3-4K^2+3K+2=0

We add all the numbers together, and all the variables

-4K^2+3K=0

a = -4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-4)·0
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{-(3)-3}{2*-4}=\frac{-6}{-8}
=3/4
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-4}=\frac{0}{-8}
=0
$