To find the roots (or solutions) of the quadratic equation $6x^2 - 19x - 7 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $6x^2 - 19x - 7 = 0$, the coefficients are:

• $a = 6$
• $b = -19$
• $c = -7$

Step 1: Calculate the discriminant

$\Delta = b^2 - 4ac = (-19)^2 - 4(6)(-7)$ $\Delta = 361 + 168 = 529$

Step 2: Use the quadratic formula

$x = \frac{-(-19) \pm \sqrt{529}}{2(6)}$ $x = \frac{19 \pm 23}{12}$

Step 3: Solve for the two roots

1. $x_1 = \frac{19 + 23}{12} = \frac{42}{12} = 3.5$
2. $x_2 = \frac{19 - 23}{12} = \frac{-4}{12} = -\frac{1}{3}$

Final solutions:

The roots of the equation are $x_1 = 3.5$ and $x_2 = -\frac{1}{3}$.

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How do you verify the roots of a quadratic equation?
2. What happens if the discriminant is negative?
3. Can all quadratic equations be factored?
4. How can we derive the quadratic formula from the general form?
5. What are the applications of quadratic equations in real life?

Tip: If the discriminant is zero, the quadratic equation has exactly one real root (a repeated root).