Question

Write the following equation in standard form. Then solve.6 q squared minus 6 qequals5 q squared minus 9 q plus 28Question content area bottomPart 1The equation in standard form isenter your response here

Answer 1

To rewrite the given equation in standard form, we will first move all terms to one side. The given equation is:

\[ 6q^2 - 6q = 5q^2 - 9q + 28 \]

Let's start by subtracting \(5q^2\) and adding \(9q\) and \(28\) to both sides:

\[
6q^2 - 6q - 5q^2 + 9q - 28 = 0
\]

Now, combine like terms:

\[
(6q^2 - 5q^2) + (-6q + 9q) - 28 = 0
\]
\[
q^2 + 3q - 28 = 0
\]

Now, the equation in standard form is:

\[ q^2 + 3q - 28 = 0 \]

Next, we can solve this quadratic equation using the quadratic formula:

\[
q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = 3\), and \(c = -28\).

Calculating the discriminant:

\[
b^2 - 4ac = 3^2 - 4(1)(-28) = 9 + 112 = 121
\]

Now substituting into the quadratic formula:

\[
q = \frac{-3 \pm \sqrt{121}}{2 \cdot 1}
\]
\[
q = \frac{-3 \pm 11}{2}
\]

This gives us two possible solutions:

1.
\[
q = \frac{-3 + 11}{2} = \frac{8}{2} = 4
\]

2.
\[
q = \frac{-3 - 11}{2} = \frac{-14}{2} = -7
\]

Thus, the solutions to the equation are:

\[
q = 4 \quad \text{and} \quad q = -7
\]