Q. Which equation shows the quadratic formula used correctly to solve \(7x^{2} = 9 + x\) for (x)?

Problem: Solve: \(7x^{2} = 9 + x\)

Step 1 – Put the equation in standard quadratic form

Start with the given equation and move all terms to one side:
\[
7x^{2} = 9 + x
\]
Subtract \(x\) and \(9\) from both sides to obtain
\[
7x^{2} – x – 9 = 0.
\]
Explanation: Moving \(x\) and \(9\) to the left changes their signs to \(-x\) and \(-9\).

Step 2 – Identify the coefficients

From
\[
7x^{2} – x – 9 = 0
\]
we read off
\[
a = 7,\quad b = -1,\quad c = -9.
\]
Explanation: \(a\) is the coefficient of \(x^{2}\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term.

Step 3 – State the quadratic formula

The solutions of \(ax^{2}+bx+c=0\) are given by
\[
x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}.
\]

Step 4 – Substitute the coefficients into the formula

Substitute \(a=7\), \(b=-1\), \(c=-9\):
\[
x = \frac{-(-1) \pm \sqrt{(-1)^{2} – 4(7)(-9)}}{2(7)}.
\]
Explanation: Note the double negative in \(-(-1)\).

Step 5 – Simplify step by step

Compute the parts:
\[
-(-1)=1,
\]
\[
(-1)^{2} – 4(7)(-9) = 1 + 252 = 253,
\]
and \(2(7)=14\). Thus
\[
x = \frac{1 \pm \sqrt{253}}{14}.
\]

Answer

The two solutions are
\[
x = \frac{1 + \sqrt{253}}{14}\quad\text{and}\quad x = \frac{1 – \sqrt{253}}{14}.
\]