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:
\[ q = \frac{-3 + 11}{2} = \frac{8}{2} = 4 \]
\[ q = \frac{-3 - 11}{2} = \frac{-14}{2} = -7 \]
Thus, the solutions to the equation are:
\[ q = 4 \quad \text{and} \quad q = -7 \]
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