If $f(x) = 2x^2 - 3x + 5$, find the value of x when $f(x)$ equals 12.

Given the function $f(x) = 2x^2 – 3x + 5$, we need to find the value of $x$ when $f(x) = 12$.

We start by setting $f(x) = 12$:

$$ 2x^2 – 3x + 5 = 12 $$

Next, we simplify the equation by subtracting 12 from both sides:

$$ 2x^2 – 3x – 7 = 0 $$

Now, we solve this quadratic equation using the quadratic formula:

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

For the equation $2x^2 – 3x – 7 = 0$, we have $a = 2$, $b = -3$, and $c = -7$. Substituting these values into the quadratic formula:

$$ x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} $$

$$ x = \frac{3 \pm \sqrt{9 + 56}}{4} $$

$$ x = \frac{3 \pm \sqrt{65}}{4} $$

Thus, the two possible solutions are:

$$ x = \frac{3 + \sqrt{65}}{4} $$

or

$$ x = \frac{3 – \sqrt{65}}{4} $$

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