$$5k^{2}-9k+18-4k^{2}=0$$
Combine $5k^{2}$ and $-4k^{2}$ to get $k^{2}$.$$k^{2}-9k+18=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $k^{2}+ak+bk+18$. To find $a$ and $b$, set up a system to be solved.$$a+b=-9$$ $$ab=1\times 18=18$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $18$.$$-1,-18$$ $$-2,-9$$ $$-3,-6$$
Calculate the sum for each pair.$$-1-18=-19$$ $$-2-9=-11$$ $$-3-6=-9$$
The solution is the pair that gives sum $-9$.$$a=-6$$ $$b=-3$$
Rewrite $k^{2}-9k+18$ as $\left(k^{2}-6k\right)+\left(-3k+18\right)$.$$\left(k^{2}-6k\right)+\left(-3k+18\right)$$
Factor out $k$ in the first and $-3$ in the second group.$$k\left(k-6\right)-3\left(k-6\right)$$
Factor out common term $k-6$ by using distributive property.$$\left(k-6\right)\left(k-3\right)$$
To find equation solutions, solve $k-6=0$ and $k-3=0$.$$k=6$$ $$k=3$$