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 will apply the quadratic formula.

First, identify the coefficients:
$a = 3$, $b = 10$, $c = -8$.

Calculate the discriminant $\Delta$:
$\Delta = b^2 - 4ac = 10^2 - 4 \times 3 \times (-8) = 100 + 96 = 196$

Since the discriminant $\Delta = 196$ is greater than zero, the quadratic equation has two distinct real roots.

Now apply the quadratic formula:
$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-10 \pm \sqrt{196}}{6}$

Calculate the roots:

• Root $x_1$:
  $x_1 = \frac{-10 + 14}{6} = \frac{4}{6} = \frac{2}{3}$
• Root $x_2$:
  $x_2 = \frac{-10 - 14}{6} = \frac{-24}{6} = -4$

Thus, the solutions are $x_1 = \frac{2}{3}$ and $x_2 = -4$.

Therefore, the solution to the equation $3x^2 + 10x - 8 = 0$ is $x_1 = \frac{2}{3}, x_2 = -4$.