4x^2+8x-21=0

1 answer

To solve the quadratic equation \(4x^2 + 8x - 21 = 0\), you can use the quadratic formula:

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

In this equation, \(a = 4\), \(b = 8\), and \(c = -21\).

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

\[ b^2 - 4ac = 8^2 - 4 \cdot 4 \cdot (-21) \]

\[ = 64 + 336 \]

\[ = 400 \]

Now, substitute the values into the quadratic formula:

\[ x = \frac{-8 \pm \sqrt{400}}{2 \cdot 4} \]

\[ = \frac{-8 \pm 20}{8} \]

Now, solve for the two possible values of \(x\):

  1. \(x = \frac{-8 + 20}{8} = \frac{12}{8} = \frac{3}{2}\)
  2. \(x = \frac{-8 - 20}{8} = \frac{-28}{8} = -\frac{7}{2}\)

So the solutions to the equation \(4x^2 + 8x - 21 = 0\) are:

\[ x = \frac{3}{2} \quad \text{and} \quad x = -\frac{7}{2} \]