The equation you provided, $2x+x-12=0$, simplifies to $3x-12=0$. This is a linear equation, not a quadratic equation. The quadratic formula is used for equations of the form $ax^2+bx+c=0$, where $a \neq 0$. However, based on our previous conversation for question 21, it's possible you meant to ask about the quadratic equation $x^2+x-12=0$. I will solve this equation using the quadratic formula. Solving $x^2+x-12=0$ using the quadratic formula: Step 1: Identify the coefficients. For a quadratic equation in the form $ax^2+bx+c=0$, we have: $a = 1$ $b = 1$ $c = -12$ Step 2: Apply the quadratic formula. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Substitute the values of $a$, $b$, and $c$: $$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(-12)}}{2(1)}$$ Step 3: Simplify the expression under the square root. $$x = \frac{-1 \pm \sqrt{1 - (-48)}}{2}$$ $$x = \frac{-1 \pm \sqrt{1 + 48}}{2}$$ $$x = \frac{-1 \pm \sqrt{49}}{2}$$ Step 4: Calculate the square root. $$x = \frac{-1 \pm 7}{2}$$ Step 5: Find the two possible values for $x$. For the positive root: $$x_1 = \frac{-1 + 7}{2}$$ $$x_1 = \frac{6}{2}$$ $$x_1 = 3$$ For the negative root: $$x_2 = \frac{-1 - 7}{2}$$ $$x_2 = \frac{-8}{2}$$ $$x_2 = -4$$ The possible values of $x$ are $3$ and $-4$. The final answer is $\boxed{3, -4}$. That's 2 down. 3 left today — send the next one.