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

Mar 18, 2017

1) Rearrange equation so that it is in the form $a {x}^{2} + b x + c = 0$
2) (Learn by heart and) use the quadratic formula to solve the equation

${x}_{1} = - \frac{2}{3}$ and ${x}_{2} = - \frac{5}{3}$

Explanation:

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

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

This is a quadratic equation of the form
$a {x}^{2} + b x + c = 0$ 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:
${x}_{1 , 2} = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
It says ${x}_{1 , 2}$ and there is a $\setminus \pm$(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:
${x}_{1 , 2} = \frac{- 21 \setminus \pm \sqrt{{21}^{2} - 4 \cdot 9 \cdot 10}}{2 \cdot 9}$
i.e.
${x}_{1 , 2} = \frac{- 21 \setminus \pm \sqrt{441 - 360}}{18}$
i.e.
${x}_{1 , 2} = \frac{- 21 \setminus \pm \sqrt{81}}{18}$
i.e.
${x}_{1 , 2} = \frac{- 21 \setminus \pm 9}{18}$
so
${x}_{1} = - \frac{12}{18}$ and ${x}_{2} = - \frac{30}{18}$
i.e.
${x}_{1} = - \frac{2}{3}$ and ${x}_{2} = - \frac{5}{3}$

Q.E.D.