Answer
$x_1 = 5;$ $x_2 = -3$
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$x^2-2x-15$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$1x^2-2x-15$
$a = 1, b = -2, c = -15$
$x = \frac{-(-2) \pm \sqrt {(-2)^2- 4(1 \times -15)}}{2(1)}$
$x = \frac{2 \pm \sqrt {4 + 60}}{2}$
$x = \frac{2 \pm \sqrt {64}}{2}$
$x_1 = \frac{2 + 8 }{2}, x_2 = \frac{2-8}{2}$
$x_1 = \frac{10}{2}, x_2 = \frac{-6}{2}$
$x_1 = 5;$ $x_2 = -3$
