## Elementary Algebra

We use the quadratic formula, which states that the solution for an equation in the form $ax^2 +bx +c =0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Using this, we obtain: a) $x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-4)}}{2} = -.6, 6.6$ b) $x = \frac{8 \pm \sqrt{8^2 - 4(4)(1)}}{2} = .5, 7.5$ c) $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-4)}}{2} = -4.8, .8$ d) $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2}=1.4, -3.4$ e) $x = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(-2)}}{2} = -.1, 14.1$ f) $x = \frac{-12 \pm \sqrt{12^2 - 4(1)(-1)}}{2} = -12.1, .1$