1. **State the problem:** Solve the quadratic equation $6q^2 - 4q = 5q^2 - 7q + 54$. 2. **Rewrite the equation in standard form:** Move all terms to one side by subtracting $5q^2 - 7q + 54$ from both sides: $$6q^2 - 4q - (5q^2 - 7q + 54) = 0$$ 3. **Simplify inside the parentheses:** $$6q^2 - 4q - 5q^2 + 7q - 54 = 0$$ 4. **Combine like terms:** $$ (6q^2 - 5q^2) + (-4q + 7q) - 54 = 0$$ $$ q^2 + 3q - 54 = 0$$ This is the quadratic equation in standard form. 5. **Use the quadratic formula:** For $aq^2 + bq + c = 0$, the solutions are $$q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=3$, and $c=-54$. 6. **Calculate the discriminant:** $$b^2 - 4ac = 3^2 - 4(1)(-54) = 9 + 216 = 225$$ 7. **Find the square root of the discriminant:** $$\sqrt{225} = 15$$ 8. **Calculate the two solutions:** $$q = \frac{-3 \pm 15}{2}$$ 9. **First solution:** $$q = \frac{-3 + 15}{2} = \frac{12}{2} = 6$$ 10. **Second solution:** $$q = \frac{-3 - 15}{2} = \frac{-18}{2} = -9$$ **Final answer:** The solutions to the equation are $q = 6$ and $q = -9$. The equation in standard form is $q^2 + 3q - 54 = 0$.