The problem asks us to solve the quadratic equation $-7f^2 + 11f + 9 = 4 - 3f$ and provide the solutions to two decimal places.

1. Problem Description

The problem asks us to solve the quadratic equation $-7f^2 + 11f + 9 = 4 - 3f$ and provide the solutions to two decimal places.

2. Solution Steps

First, we need to rewrite the equation in the standard quadratic form, $ax^2 + bx + c = 0$. In this case, the variable is $f$.

$-7f^2 + 11f + 9 = 4 - 3f$

Add $3f$ to both sides:

$-7f^2 + 11f + 3f + 9 = 4 - 3f + 3f$

$-7f^2 + 14f + 9 = 4$

Subtract 4 from both sides:

$-7f^2 + 14f + 9 - 4 = 4 - 4$

$-7f^2 + 14f + 5 = 0$

Now we have a quadratic equation in the form $af^2 + bf + c = 0$, where $a = -7$, $b = 14$, and $c = 5$.

We can solve for $f$ using the quadratic formula:

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

Substituting the values of $a$, $b$, and $c$:

$f = \frac{-14 \pm \sqrt{14^2 - 4(-7)(5)}}{2(-7)}$

$f = \frac{-14 \pm \sqrt{196 + 140}}{-14}$

$f = \frac{-14 \pm \sqrt{336}}{-14}$

$f = \frac{-14 \pm \sqrt{16 \times 21}}{-14}$

$f = \frac{-14 \pm 4\sqrt{21}}{-14}$

$f = \frac{14 \mp 4\sqrt{21}}{14}$

$f = 1 \mp \frac{2\sqrt{21}}{7}$

Now we find the two solutions:

$f_1 = 1 + \frac{2\sqrt{21}}{7} \approx 1 + \frac{2(4.58)}{7} \approx 1 + \frac{9.16}{7} \approx 1 + 1.3086 \approx 2.31$

$f_2 = 1 - \frac{2\sqrt{21}}{7} \approx 1 - \frac{2(4.58)}{7} \approx 1 - \frac{9.16}{7} \approx 1 - 1.3086 \approx -0.31$

So the two solutions, rounded to two decimal places, are $2.31$ and $-0.31$.

3. Final Answer

The solutions to the quadratic equation are $f = 2.31$ and $f = -0.31$.