Solve -n(n-3)+1=4n-1

$$- n ( n - 3 ) + 1 = 4 n - 1$$

Solve for n

$$n=-2$$

$$n=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -n by n-3.

$$\left(-n\right)n-3\left(-n\right)+1=4n-1$$

Multiply -3 and -1 to get 3.

$$\left(-n\right)n+3n+1=4n-1$$

Subtract 4n from both sides.

$$\left(-n\right)n+3n+1-4n=-1$$

Combine 3n and -4n to get -n.

$$\left(-n\right)n-n+1=-1$$

Add 1 to both sides.

$$\left(-n\right)n-n+1+1=0$$

Add 1 and 1 to get 2.

$$\left(-n\right)n-n+2=0$$

Multiply n and n to get n^{2}.

$$-n^{2}-n+2=0$$

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

$$n=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-\left(-1\right)±\sqrt{1+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$n=\frac{-\left(-1\right)±\sqrt{1+8}}{2\left(-1\right)}$$

Add 1 to 8.

$$n=\frac{-\left(-1\right)±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$n=\frac{-\left(-1\right)±3}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$n=\frac{1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{1±3}{-2}$$

Now solve the equation n=\frac{1±3}{-2} when ± is plus. Add 1 to 3.

$$n=\frac{4}{-2}$$

Divide 4 by -2.

$$n=-2$$

Now solve the equation n=\frac{1±3}{-2} when ± is minus. Subtract 3 from 1.

$$n=-\frac{2}{-2}$$

Divide -2 by -2.

$$n=1$$

The equation is now solved.

$$n=-2 n=1$$

Steps for Completing the Square

Use the distributive property to multiply -n by n-3.

$$\left(-n\right)n-3\left(-n\right)+1=4n-1$$

Multiply -3 and -1 to get 3.

$$\left(-n\right)n+3n+1=4n-1$$

Subtract 4n from both sides.

$$\left(-n\right)n+3n+1-4n=-1$$

Combine 3n and -4n to get -n.

$$\left(-n\right)n-n+1=-1$$

Subtract 1 from both sides.

$$\left(-n\right)n-n=-1-1$$

Subtract 1 from -1 to get -2.

$$\left(-n\right)n-n=-2$$

Multiply n and n to get n^{2}.

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

Divide both sides by -1.

$$\frac{-n^{2}-n}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\left(-\frac{1}{-1}\right)n=-\frac{2}{-1}$$

Divide -1 by -1.

$$n^{2}+n=-\frac{2}{-1}$$

Divide -2 by -1.

$$n^{2}+n=2$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=2+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+n+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{9}{4}$$

Factor n^{2}+n+\frac{1}{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}.

$$\left(n+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{3}{2} n+\frac{1}{2}=-\frac{3}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$n=1 n=-2$$