+397 This quadratic, \(z^2 - z - 3 = 0\) is a bit trickier. For this one, we need to use the quadratic formula, which is \(\frac{-b ± \sqrt{b^2 - 4ac}}{2a}\).
When we plug in the values of \(a, b,\) and \(c\), we get \(\frac{-(-1) ± \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1±\sqrt{13}}{2}\). This means \(z = \frac{1}{2} + \frac{1}{2}\sqrt{13}\) or \(z = \frac{1}{2} + \frac{-1}{2}\sqrt{13}\).
- Daisy
+397 This quadratic, \(z^2 - z - 3 = 0\) is a bit trickier. For this one, we need to use the quadratic formula, which is \(\frac{-b ± \sqrt{b^2 - 4ac}}{2a}\).
When we plug in the values of \(a, b,\) and \(c\), we get \(\frac{-(-1) ± \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1±\sqrt{13}}{2}\). This means \(z = \frac{1}{2} + \frac{1}{2}\sqrt{13}\) or \(z = \frac{1}{2} + \frac{-1}{2}\sqrt{13}\).
- Daisy
