����ý

Solve \(x^2 + 6x - 12 = 0\) using the quadratic formula.

First, identify the value of \(a\), \(b\) and \(c\). In this example, \(a = 1\), \(b = 6\) and \(c = -12\).

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(\frac{(- 6) \pm \sqrt{6^2 - 4 \times 1 \times -12}}{2 \times 1}\)

\(\frac{(-6) \pm \sqrt{36 + 48}}{2}\)

\(\frac{(-6) \pm \sqrt{84}}{2}\)

\(\frac{(-6) \pm \sqrt{84}}{2}\)

The first solution is \(\frac{(-6) + \sqrt{84}}{2} = 1.58\) (2 dp).

The second solution is \(\frac{(-6) - \sqrt{84}}{2} = -7.58\) (2 dp).

\(x^2 + 2x + 5 = 0\)

Using the quadratic formula to solve this equation with \(a = 1\), \(b = 2\) and \(c = 5\) gives:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times 5}}{2 \times 1} = \frac{-2\pm \sqrt{-16}}{2}\)

The graph of \(y = x^2 + 2x + 5\) does not cross or touch the x-axis as the equation \(x^2 + 2x + 5 = 0\) has no roots.