Question

How do you solve `9x^2+11x+18=-10x+8`?

Answer 1

1) Rearrange equation so that it is in the form `ax^2 + bx +c =0`
2) (Learn by heart and) use the quadratic formula to solve the equation

Answers are, in this case,
`x_1 = -2/3` and `x_2 = -5/3`

Explanation:

Ohkay, let's do this.
1) Rearrange every term on one side, say to the left.
We now have:
`9x^2 + 11x + 18 +10x -8 = 0`
(the sign-change occurs because the terms "crossed" the equal sign)

Now, you can simplify a bit. We have:
`9x^2 + 21x +10 = 0`

This is a quadratic equation of the form
`ax^2 + bx + c =0` where a,b,c are real numbers (in our case, they are integers, which are also real numbers but enough side-tracking).
To solve this, you need the quadratic formula which is, in its general form:
`x_(1,2) = (-b \pm sqrt(b^2 - 4ac))/(2a)`
It says `x_(1,2)` and there is a `\pm`(plus or minus) sign in this equation because there can be a maximum of two (real) solutions, `x_1` and `x_2` which depend on whether we use the plus sign or the minus sign in that equation.

Anyway, applying this formula to our equation, we need to put `a = 9`, `b=21`, and `c=10`
we get the following:
`x_(1,2) = (-21 \pm sqrt(21^2 - 4*9*10))/(2*9)`
i.e.
`x_(1,2) = (-21 \pm sqrt(441 - 360))/(18)`
i.e.
`x_(1,2) = (-21 \pm sqrt(81))/(18)`
i.e.
`x_(1,2) = (-21 \pm 9)/(18)`
so
`x_1 = -12/18` and `x_2 = -30/18`
i.e.
`x_1 = -2/3` and `x_2 = -5/3`

Q.E.D.