Solve each quadratic equation using the method that seems most appropriate. 2 n 2 $-$ 6 n = 3
A) $n = \frac { 3 + \sqrt { 3 } } { 2 } , n = \frac { 3 - \sqrt { 3 } } { 2 }$
B) $n = \frac { 6 + \sqrt { 3 } } { 2 } , n = \frac { - 6 - \sqrt { 3 } } { 2 }$
C) $n = \frac { - 6 + \sqrt { 3 } } { 2 } , n = \frac { 6 + \sqrt { 3 } } { 2 }$

Question

Quizplus · Accepted Answer

This equation can be solved using either factoring or the quadratic formula. However, factoring may be difficult since there is no common factor, and the quadratic formula is more straightforward. The quadratic formula is:

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

Plugging in the values from the given equation, we have:

$$n = \frac{-(-6)\pm\sqrt{(-6)^2-4(2)(-3)}}{2(2)}$$

Simplifying, we get:

$$n = \frac{6\pm\sqrt{60}}{4}$$

Further simplifying, we get:

$$n = \frac{3\pm\sqrt{15}}{2}$$

Thus, the answer is choice A.

Solve Each Quadratic Equation Using the Method That Seems Most −-−

Solve Each Quadratic Equation Using the Method That Seems Most $-$