6x^2-9x+18
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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

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116