Question

Solve the following equation for t. −2t² + t + 28 = 0

A. t = 2/7, −4
B. t = 4/3, −7
C. t = −7/2, 4
D. t = −3/4, 7

Answer 1

The quadratic equation −2t² + t + 28 = 0 is solved using the quadratic formula, yielding t = -7/2 and t = 4 as the solutions, corresponding to option C.

Step-by-step explanation:

The quadratic equation −2t² + t + 28 = 0 is solved using the quadratic formula, yielding t = -7/2 and t = 4 as the solutions, corresponding to option C.

To solve for t in the quadratic equation −2t² + t + 28 = 0, we'll use the quadratic formula, which is -b ± √b² - 4ac / (2a), where a, b, and c represent the coefficients in the equation at² + bt + c = 0.

In our case, a = -2, b = 1, and c = 28. Plugging these values into the quadratic formula gives us:

t = (-1 ± √(1² - 4(-2)(28))) / (2(-2))
t = (-1 ± √(1 + 224)) / (-4)
t = (-1 ± √225) / (-4)
t = (-1 ± 15) / (-4)

Therefore, we have two possible solutions for t:

t = (-1 + 15) / (-4) = 14 / (-4) = -7/2 or t = 3.5
t = (-1 - 15) / (-4) = -16 / (-4) = 4

So, the solutions are t = -7/2, 4, which corresponds to option C in the multiple-choice answers.