Answer
$x_1 = 7, x_2 = 6$
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$x^2-13x+42$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$1x^2-13x+42$
$a = 1, b = -13, c = 42$
$x = \frac{-(-13) \pm \sqrt {(-13)^2- 4(1 \times 42)}}{2(1)}$
$x = \frac{13 \pm \sqrt {169- 168}}{2}$
$x = \frac{13 \pm \sqrt {1}}{2}$
$x_1 = \frac{13+1}{2}, x_2 = \frac{13-1}{2}$
$x_1 = \frac{14}{2}, x_2 = \frac{12}{2}$
$x_1 = 7, x_2 = 6$
