To rewrite the equation \( 6q^2 - 5q = 5q^2 - 7q + 48 \) in standard form, we first move all terms to one side of the equation.
Start with the original equation:
\[ 6q^2 - 5q = 5q^2 - 7q + 48 \]
Subtract \( 5q^2 \), add \( 7q \), and subtract \( 48 \) from both sides:
\[ 6q^2 - 5q - 5q^2 + 7q - 48 = 0 \]
Now, combine like terms:
\[ (6q^2 - 5q^2) + (-5q + 7q) - 48 = 0 \] \[ q^2 + 2q - 48 = 0 \]
This is the equation in standard form.
Now, we can solve the quadratic equation \( q^2 + 2q - 48 = 0 \) using the quadratic formula:
\[ q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2 \), and \( c = -48 \).
First, calculate the discriminant:
\[ b^2 - 4ac = 2^2 - 4(1)(-48) = 4 + 192 = 196 \]
Now apply the quadratic formula:
\[ q = \frac{-2 \pm \sqrt{196}}{2(1)} \] \[ q = \frac{-2 \pm 14}{2} \]
Now, we find the two possible values for \( q \):
\[ q = \frac{-2 + 14}{2} = \frac{12}{2} = 6 \]
\[ q = \frac{-2 - 14}{2} = \frac{-16}{2} = -8 \]
Thus, the solutions of the equation \( q^2 + 2q - 48 = 0 \) are:
Answer:
The equation in standard form is \( q^2 + 2q - 48 = 0 \).
The solutions are \( q = 6 \) and \( q = -8 \).