The problem is to solve the quadratic equation $9x^2 + 7 = 18$ using the quadratic formula.

1. Problem Description

The problem is to solve the quadratic equation $9x^2 + 7 = 18$ using the quadratic formula.

2. Solution Steps

First, rewrite the equation in the standard form $ax^2 + bx + c = 0$.

$9x^2 + 7 = 18$

$9x^2 + 7 - 18 = 0$

$9x^2 - 11 = 0$

Here, $a = 9$, $b = 0$, and $c = -11$.

Now, use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$ into the formula:

$x = \frac{-0 \pm \sqrt{0^2 - 4(9)(-11)}}{2(9)}$

$x = \frac{\pm \sqrt{0 + 396}}{18}$

$x = \frac{\pm \sqrt{396}}{18}$

Simplify the square root:

$\sqrt{396} = \sqrt{36 \cdot 11} = \sqrt{36} \cdot \sqrt{11} = 6\sqrt{11}$

So, $x = \frac{\pm 6\sqrt{11}}{18}$

$x = \frac{\pm \sqrt{11}}{3}$

Therefore, the two solutions are $x = \frac{\sqrt{11}}{3}$ and $x = -\frac{\sqrt{11}}{3}$.

3. Final Answer

$\frac{\sqrt{11}}{3}, -\frac{\sqrt{11}}{3}$