Question

2x^2-3x-15=-4x

Use the quadric formula to solve. Express your answer in simplest form.

Answer 1

To solve the equation \(2x^2 - 3x - 15 = -4x\), we first move all terms to one side of the equation:

\[
2x^2 - 3x + 4x - 15 = 0
\]

This simplifies to:

\[
2x^2 + x - 15 = 0
\]

Now, we can use the quadratic formula to solve for \(x\). The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation \(2x^2 + x - 15 = 0\), we identify the coefficients as follows:

- \(a = 2\)
- \(b = 1\)
- \(c = -15\)

Next, we calculate the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-15)
\]
\[
= 1 + 120
\]
\[
= 121
\]

Now we can plug \(a\), \(b\), and the discriminant into the quadratic formula:

\[
x = \frac{-1 \pm \sqrt{121}}{2 \cdot 2}
\]
\[
= \frac{-1 \pm 11}{4}
\]

This gives us two potential solutions:

1. \(x = \frac{-1 + 11}{4} = \frac{10}{4} = \frac{5}{2}\)

2. \(x = \frac{-1 - 11}{4} = \frac{-12}{4} = -3\)

Thus, the solutions to the equation \(2x^2 - 3x - 15 = -4x\) are:

\[
\boxed{\frac{5}{2}} \quad \text{and} \quad \boxed{-3}
\]