Solve the Quadratic Equation: 3x²+10x-8=0

Solve the following equation:

$3x^2+10x-8=0$

To solve the quadratic equation $3x^2+10x-8=0$, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = 10$, and $c = -8$.

First, calculate the discriminant $b^2 - 4ac$:
$b^2 = 10^2 = 100$
$4ac = 4 \times 3 \times (-8) = -96$
Thus, $b^2 - 4ac = 100 + 96 = 196$.

Since the discriminant (196) is positive, the equation has two distinct real solutions.

Now, substitute into the quadratic formula:
$x = \frac{-10 \pm \sqrt{196}}{2 \times 3} = \frac{-10 \pm 14}{6}$.

This results in two solutions:

• For the plus sign: $x_1 = \frac{-10 + 14}{6} = \frac{4}{6} = \frac{2}{3}$
• For the minus sign: $x_2 = \frac{-10 - 14}{6} = \frac{-24}{6} = -4$

Therefore, the solutions to the given quadratic equation are $x_1 = \frac{2}{3}$ and $x_2 = -4$.

Comparing these results with the multiple-choice options provided, the correct answer is choice 3: $x_1 = \frac{2}{3}, x_2 = -4$.