Absolutely! Let's solve the given quadratic equation step-by-step.
The equation provided is:
[tex]\[
-v^2 + 5v + 6 = -2v^2
\][/tex]
To solve this, we first need to move all the terms to one side to set the equation equal to zero. We'll start by adding [tex]\(2v^2\)[/tex] to both sides:
[tex]\[
-v^2 + 5v + 6 + 2v^2 = -2v^2 + 2v^2
\][/tex]
This simplifies to:
[tex]\[
v^2 + 5v + 6 = 0
\][/tex]
Now, we have a standard quadratic equation in the form [tex]\(av^2 + bv + c = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 6\)[/tex].
Next, we use the quadratic formula to solve for [tex]\(v\)[/tex]. The quadratic formula is given by:
[tex]\[
v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
First, we calculate the discriminant, [tex]\(b^2 - 4ac\)[/tex]:
[tex]\[
b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1
\][/tex]
Now that we have the discriminant, we substitute back into the quadratic formula:
[tex]\[
v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{1}}{2 \cdot 1} = \frac{-5 \pm 1}{2}
\][/tex]
This gives us two possible solutions:
1. When we use the positive root:
[tex]\[
v = \frac{-5 + 1}{2} = \frac{-4}{2} = -2
\][/tex]
2. When we use the negative root:
[tex]\[
v = \frac{-5 - 1}{2} = \frac{-6}{2} = -3
\][/tex]
Therefore, the solutions to the quadratic equation are:
[tex]\[
v = -2 \quad \text{and} \quad v = -3
\][/tex]
So, the values of [tex]\(v\)[/tex] that satisfy the equation [tex]\( -v^2 + 5v + 6 = -2v^2 \)[/tex] are [tex]\(v = -2\)[/tex] and [tex]\(v = -3\)[/tex].
