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