How do you solve #2x^2-x-4=10#?

Question

Avery Alexander · Accepted Answer

#(1 +- sqrt113)/4# Use the improved quadratic formula (Socratic Search). #y = 2x^2 - x - 14 = 0# #D = d^2 = b^2 - 4ac = 1 + 112 = 113# --> #d = +- sqrt113# There are 2 real roots: #x = -b/(2a) +- d/(2a) = 1/4 +- sqrt113/4 = (1 + sqrt113)/4#

Jameson Allen · Answer

undefined To solve the equation 2x^2 - x - 4 = 10, follow these steps:

1. Subtract 10 from both sides to move the constant to the right side:
   2x^2 - x - 4 - 10 = 0
   2x^2 - x - 14 = 0

2. To factor the quadratic equation, find two numbers that multiply to -28 (2 * -14) and add up to -1 (coefficient of x):
   -4 and 3 satisfy these conditions because -4 * 3 = -12 and -4 + 3 = -1
   So, rewrite -x as -4x + 3x:
   2x^2 - 4x + 3x - 14 = 0

3. Factor by grouping:
   2x(x - 2) + 3(x - 2) = 0
   (2x + 3)(x - 2) = 0

4. Set each factor to zero and solve for x:
   2x + 3 = 0 or x - 2 = 0
   2x = -3 or x = 2
   x = -3/2 or x = 2

So, the solutions to the equation are x = -3/2 or x = 2.

Savannah Anderson · Answer

undefined To solve the equation $2x^2 - x - 4 = 10$, follow these steps:

1. Move all terms to one side of the equation to set it equal to zero:
$$2x^2 - x - 4 - 10 = 0$$

2. Combine like terms:
$$2x^2 - x - 14 = 0$$

3. Now, we have a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 2$, $b = -1$, and $c = -14$. To solve this quadratic equation, you can use the quadratic formula:
$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

4. Substitute the values of $a$, $b$, and $c$ into the quadratic formula:
$$x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(2)(-14)}}}}{{2(2)}}$$

5. Simplify inside the square root:
$$x = \frac{{1 \pm \sqrt{{1 + 112}}}}{{4}}$$
$$x = \frac{{1 \pm \sqrt{{113}}}}{{4}}$$

So, the solutions to the equation $2x^2 - x - 4 = 10$ are:
$$x = \frac{{1 + \sqrt{{113}}}}{{4}} \text{ or } x = \frac{{1 - \sqrt{{113}}}}{{4}}$$