Answer
$x_1 = -1, x_2 = -\frac{4}{3}$
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$3x^2+7x+4$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$3x^2+7x+4$
$a = 3, b = 7, c = 4$
$x = \frac{-(7) \pm \sqrt {(7)^2- 4(3 \times 4)}}{2(3)}$
$x = \frac{-7 \pm \sqrt {49- 48}}{6}$
$x = \frac{-7 \pm \sqrt {1}}{6}$
$x_1 = \frac{-7+1}{6}, x_2 = \frac{-7-1}{6}$
$x_1 = \frac{-6}{6}, x_2 = \frac{-8}{6}$
$x_1 = -1, x_2 = -\frac{4}{3}$
