The problem asks us to find the values of $x$ for which the quadratic function $f(x) = x^2 + 7x + 7$ is greater than 0. We are instructed to use the quadratic formula and round our answer to two decimal places.

1. Problem Description

The problem asks us to find the values of $x$ for which the quadratic function $f(x) = x^2 + 7x + 7$ is greater than

0. We are instructed to use the quadratic formula and round our answer to two decimal places.

2. Solution Steps

First, we need to find the roots of the quadratic equation $x^2 + 7x + 7 = 0$. We can use the quadratic formula:

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

In our case, $a = 1$, $b = 7$, and $c = 7$. Plugging these values into the quadratic formula, we get:

$x = \frac{-7 \pm \sqrt{7^2 - 4(1)(7)}}{2(1)}$

$x = \frac{-7 \pm \sqrt{49 - 28}}{2}$

$x = \frac{-7 \pm \sqrt{21}}{2}$

Now, we calculate the two roots:

$x_1 = \frac{-7 + \sqrt{21}}{2} \approx \frac{-7 + 4.58}{2} \approx \frac{-2.42}{2} \approx -1.21$

$x_2 = \frac{-7 - \sqrt{21}}{2} \approx \frac{-7 - 4.58}{2} \approx \frac{-11.58}{2} \approx -5.79$

So the roots are approximately $x = -1.21$ and $x = -5.79$.

Since the coefficient of the $x^2$ term is positive (1), the parabola opens upwards. Therefore, the function $f(x) = x^2 + 7x + 7$ is greater than 0 when $x$ is less than the smaller root and when $x$ is greater than the larger root. In other words, $f(x) > 0$ when $x < -5.79$ or $x > -1.21$.

3. Final Answer

$x < -5.79, x > -1.21$