2x^2-2x-12=0 , what are the two equations?

To solve the quadratic equation 2x^2-2x-12=0, the quadratic formula is used yielding two solutions, x = 3 and x = -2.

To find the solutions for the quadratic equation 2x^2-2x-12=0, we can use the quadratic formula. The general form of a quadratic equation is ax^2 + bx + c = 0. For the given equation, the coefficients are a = 2, b = -2, and c = -12. Applying the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a), we substitute in our coefficients to get:

x = (2 ± √((-2)^2 - 4(2)(-12))) / (2(2))

x = (2 ± √(4 + 96)) / 4

x = (2 ± √(100)) / 4

x = (2 ± 10) / 4

Thus, the two solutions for x are:

Therefore, the two solutions of the quadratic equation 2x^2-2x-12=0 are x = 3 and x = -2.

To solve the equation 2x^2 - 2x - 12 = 0, we can use the quadratic formula. The solutions are x = 3 and x = -2.

To solve the equation 2x^2 - 2x - 12 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

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

For our equation, a = 2, b = -2, and c = -12. Plugging in these values, we get:

$$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-12)}}{2(2)}$$

Calculating the values inside the square root:

$$x = \frac{2 \pm \sqrt{4 + 96}}{4}$$

$$x = \frac{2 \pm \sqrt{100}}{4}$$

$$x = \frac{2 \pm 10}{4}$$

So the two solutions to the equation are:

$$x = \frac{2 + 10}{4}$$

$$x = \frac{12}{4}$$

$$x = 3$$

and

$$x = \frac{2 - 10}{4}$$

$$x = \frac{-8}{4}$$

$$x = -2$$

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