Answer
$x_1 = 1, x_2 = -\frac{3}{2}$
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$2x^2+x-3$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$2x^2+1x-3$
$a = 2, b = 1, c = 3$
$x = \frac{-(1) \pm \sqrt {(1)^2- 4(2 \times -3)}}{2(2)}$
$x = \frac{-1 \pm \sqrt {1+24}}{4}$
$x = \frac{-1 \pm \sqrt {25}}{4}$
$x_1 = \frac{-1+5}{4}, x_2 = \frac{-1 - 5}{4}$
$x_1 = \frac{4}{4}, x_2 = \frac{-6}{4}$
$x_1 = 1, x_2 = -\frac{3}{2}$
