K(h)=h^2-4h+3

Solution for K(h)=h^2-4h+3 equation:

(K)=K^2-4K+3

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1K^2+5K-3=0

a = -1; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·(-1)·(-3)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{13}}{2*-1}=\frac{-5-\sqrt{13}}{-2}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*-1}=\frac{-5+\sqrt{13}}{-2}
$