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lys-0071 [83]
2 years ago
7

Solve using the quadratic formula 2n^2- 15 = -4n

Mathematics
1 answer:
NeX [460]2 years ago
6 0

Answer:

\displaystyle n_1 = \frac{-2 + \sqrt{34}}{2} \approx 1.9155 \text{ or } n_2 = \frac{-2-\sqrt{34}}{2} \approx -3.9155

Step-by-step explanation:

We are given the equation:

\displaystyle 2n^2 - 15 = -4n

And we want to solve using the quadratic formula.

First, isolate the equation:

\displaystyle 2n^2 + 4n - 15 = 0

Recall that for equations in the form <em>ax</em>² + <em>bx</em> + <em>c</em> = 0, the solutions are given by:

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

In this case, <em>a</em> = 2, <em>b</em> = 4, and <em>c</em> = -15.

Substitute and evaluate:

\displaystyle \begin{aligned} n&= \frac{-(4)\pm\sqrt{(4)^2-4(2)(-15)}}{2(2)} \\ \\ &= \frac{-4\pm\sqrt{136}}{4} \\ \\ &= \frac{-4\pm2\sqrt{34}}{4} \\ \\ &= \frac{-2\pm\sqrt{34}}{2} \end{aligned}

In conclusion, our two solutions are:

\displaystyle n_1 = \frac{-2 + \sqrt{34}}{2} \approx 1.9155 \text{ or } n_2 = \frac{-2-\sqrt{34}}{2} \approx -3.9155

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