Hey there! I'm happy to help!
This is a quadratic equation (it follows the form ax^2+bx+c). To solve these, we first want to factor. We have to find two numbers that multiply to create c and multiply to get b. We'll call these numbers s and t.
s·t=18
s+t=-9
Let's look at factors of 18.
18 and 1
2 and 9
3 and 6
We see that 3 and 6 add to make 9. But, we want -9. If we use -3 and -6, it will add to make -9 and still multiply to get 18.
So, our numbers are -3 and -6. We put this into the factored form (x+s)(x+t), so we have (x-3)(x-6). To find what x is equal to, we set that factored form equal to zero and we see what x values would give us 0.
(x-3)(x-6)=0
There are two possible ways to make this 0. The first parentheses end up equaling 0 and the second ones equal 0. If we have x-3=0, then x=3. If we have x-6=0, then we have x=6.
Therefore, the two solutions are x=3 or x=6.
Have a wonderful day! :D
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