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Solve 2x^(2)+x-45 to find two values of x that each give an output of 35

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Final answer:

The quadratic equation 2x^2 + x - 45 = 35 is solved by first equalizing it to zero to obtain 2x^2 + x - 80 = 0. Then, by applying the quadratic formula with a = 2, b = 1, and c = -80, we find the two values of x to be (-1 + √641) / 4 and (-1 - √641) / 4.

Step-by-step explanation:

To solve the quadratic equation 2x^2 + x - 45 = 35, we first bring all terms to one side of the equation to set it equal to zero:

2x^2 + x - 45 - 35 = 0

2x^2 + x - 80 = 0

Now we apply the quadratic formula x = (-b ± √(b^2 - 4ac)) / (2a) to find the values of x. In this equation a = 2, b = 1, and c = -80. Plugging these into the formula gives us:

x = (-(1) ± √((1)^2 - 4(2)(-80))) / (2(2))

x = (-1 ± √(1 + 640)) / 4

x = (-1 ± √641) / 4

Thus, the two possible values for x are:

x = (-1 + √641) / 4 and x = (-1 - √641) / 4.

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