Solve 94=n^2-3n+8/2

$$94 = n ^ { 2 } - 3 n + 8 / 2$$

Solve for n

$$n = \frac{3 \sqrt{41} + 3}{2} \approx 11.104686356$$

$$n=\frac{3-3\sqrt{41}}{2}\approx -8.104686356$$

Steps Using the Quadratic Formula

Divide 8 by 2 to get 4.

$$94=n^{2}-3n+4$$

Swap sides so that all variable terms are on the left hand side.

$$n^{2}-3n+4=94$$

Subtract 94 from both sides.

$$n^{2}-3n+4-94=0$$

Subtract 94 from 4 to get -90.

$$n^{2}-3n-90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-90\right)}}{2}$$

Square -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\left(-90\right)}}{2}$$

Multiply -4 times -90.

$$n=\frac{-\left(-3\right)±\sqrt{9+360}}{2}$$

Add 9 to 360.

$$n=\frac{-\left(-3\right)±\sqrt{369}}{2}$$

Take the square root of 369.

$$n=\frac{-\left(-3\right)±3\sqrt{41}}{2}$$

The opposite of -3 is 3.

$$n=\frac{3±3\sqrt{41}}{2}$$

Now solve the equation n=\frac{3±3\sqrt{41}}{2} when ± is plus. Add 3 to 3\sqrt{41}.

$$n=\frac{3\sqrt{41}+3}{2}$$

Now solve the equation n=\frac{3±3\sqrt{41}}{2} when ± is minus. Subtract 3\sqrt{41} from 3.

$$n=\frac{3-3\sqrt{41}}{2}$$

The equation is now solved.

$$n=\frac{3\sqrt{41}+3}{2} n=\frac{3-3\sqrt{41}}{2}$$

Steps for Completing the Square

Divide 8 by 2 to get 4.

$$94=n^{2}-3n+4$$

Swap sides so that all variable terms are on the left hand side.

$$n^{2}-3n+4=94$$

Subtract 4 from both sides.

$$n^{2}-3n=94-4$$

Subtract 4 from 94 to get 90.

$$n^{2}-3n=90$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=90+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-3n+\frac{9}{4}=90+\frac{9}{4}$$

Add 90 to \frac{9}{4}.

$$n^{2}-3n+\frac{9}{4}=\frac{369}{4}$$

Factor n^{2}-3n+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{3}{2}\right)^{2}=\frac{369}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{369}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{3\sqrt{41}}{2} n-\frac{3}{2}=-\frac{3\sqrt{41}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$n=\frac{3\sqrt{41}+3}{2} n=\frac{3-3\sqrt{41}}{2}$$