Questions: Solve for (z). [2 z^2-13 z+15=0] Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas. [z=]

Solution Steps

The given equation is a quadratic equation in the form: $$2z^{2} - 13z + 15 = 0$$ where $a = 2$, $b = -13$, and $c = 15$.

The quadratic formula is: $$z = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ Substitute $a = 2$, $b = -13$, and $c = 15$ into the formula: $$z = \frac{-(-13) \pm \sqrt{(-13)^{2} - 4 \cdot 2 \cdot 15}}{2 \cdot 2}$$

Calculate the discriminant: $$\Delta = b^{2} - 4ac = (-13)^{2} - 4 \cdot 2 \cdot 15 = 169 - 120 = 49$$ Since the discriminant is positive, there are two real solutions.

Substitute the discriminant back into the quadratic formula: $$z = \frac{13 \pm \sqrt{49}}{4} = \frac{13 \pm 7}{4}$$ This gives two solutions: $$z = \frac{13 + 7}{4} = \frac{20}{4} = 5$$ and $$z = \frac{13 - 7}{4} = \frac{6}{4} = \frac{3}{2}$$

Final Answer