Question Use the quadratic formula to solve Express your answer in simplest form. 8z^2+18z-7=-2 Answer Attemptiout of 2 z=square

To solve the quadratic equation $8z^2 + 18z - 7 = -2$, we first need to rewrite it in standard form $az^2 + bz + c = 0$.

Starting with:
$$8z^2 + 18z - 7 = -2$$

Add 2 to both sides to set the equation to 0:
$$8z^2 + 18z - 7 + 2 = 0$$
$$8z^2 + 18z - 5 = 0$$

Now, we have the quadratic equation in standard form where $a = 8$, $b = 18$, and $c = -5$.

Next, we use the quadratic formula:
$$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Substitute $a$, $b$, and $c$ into the formula:
$$z = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 8 \cdot (-5)}}{2 \cdot 8}$$
$$z = \frac{-18 \pm \sqrt{324 + 160}}{16}$$
$$z = \frac{-18 \pm \sqrt{484}}{16}$$
$$z = \frac{-18 \pm 22}{16}$$

This gives us two solutions:
$$z = \frac{-18 + 22}{16} = \frac{4}{16} = \frac{1}{4}$$
$$z = \frac{-18 - 22}{16} = \frac{-40}{16} = -\frac{5}{2}$$

So, the solutions are:
$$z = \frac{1}{4} \quad \text{or} \quad z = -\frac{5}{2}$$