Question

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

Answer 1

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 ax² + bx + c = 0, the solutions are given by:

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

In this case, a = 2, b = 4, and c = -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:

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