Solve for: 5 k ^ 2-9 k+18 = 4 k ^ 2

Expression: $$5 k ^ { 2 } - 9 k + 18 = 4 k ^ { 2 }$$

Subtract $4k^{2}$ from both sides.

$$5k^{2}-9k+18-4k^{2}=0$$

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 $k-3=0$.

$$k=6$$ $$k=3$$