K^3-5k^2+16k-12=0

Solution for K^3-5k^2+16k-12=0 equation:

^3-5K^2+16K-12=0

We add all the numbers together, and all the variables

-5K^2+16K=0

a = -5; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-5)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-5}=\frac{-32}{-10}
=3+1/5
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-5}=\frac{0}{-10}
=0
$