Solve -5-n/4-1*n-0/1+1=-1

$$\frac { - 5 - n } { 4 - 1 } \cdot \frac { n - 0 } { 1 + 1 } = - 1$$

Solve for n

$$n=-6$$

$$n=1$$

Steps Using the Quadratic Formula

Subtract 1 from 4 to get 3.

$$\frac{-5-n}{3}\times \frac{n-0}{1+1}=-1$$

Add 1 and 1 to get 2.

$$\frac{-5-n}{3}\times \frac{n-0}{2}=-1$$

Multiply \frac{-5-n}{3} times \frac{n-0}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(-5-n\right)\left(n-0\right)}{3\times 2}=-1$$

Multiply 3 and 2 to get 6.

$$\frac{\left(-5-n\right)\left(n-0\right)}{6}=-1$$

Add 1 to both sides.

$$\frac{\left(-5-n\right)\left(n-0\right)}{6}+1=0$$

Use the distributive property to multiply -5-n by n-0.

$$\frac{-5\left(n-0\right)-n\left(n-0\right)}{6}+1=0$$

Multiply both sides of the equation by 6.

$$-5\left(n-0\right)-n\left(n-0\right)+6=0$$

Reorder the terms.

$$-nn-5n+6=0$$

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

$$-n^{2}-5n+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square -5.

$$n=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-\left(-5\right)±\sqrt{25+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$n=\frac{-\left(-5\right)±\sqrt{25+24}}{2\left(-1\right)}$$

Add 25 to 24.

$$n=\frac{-\left(-5\right)±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$n=\frac{-\left(-5\right)±7}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$n=\frac{5±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{5±7}{-2}$$

Now solve the equation n=\frac{5±7}{-2} when ± is plus. Add 5 to 7.

$$n=\frac{12}{-2}$$

Divide 12 by -2.

$$n=-6$$

Now solve the equation n=\frac{5±7}{-2} when ± is minus. Subtract 7 from 5.

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

Divide -2 by -2.

$$n=1$$

The equation is now solved.

$$n=-6 n=1$$

Steps for Completing the Square

Subtract 1 from 4 to get 3.

$$\frac{-5-n}{3}\times \frac{n-0}{1+1}=-1$$

Add 1 and 1 to get 2.

$$\frac{-5-n}{3}\times \frac{n-0}{2}=-1$$

Multiply \frac{-5-n}{3} times \frac{n-0}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(-5-n\right)\left(n-0\right)}{3\times 2}=-1$$

Multiply 3 and 2 to get 6.

$$\frac{\left(-5-n\right)\left(n-0\right)}{6}=-1$$

Multiply both sides by 6.

$$\left(-5-n\right)\left(n-0\right)=-6$$

Reorder the terms.

$$n\left(-n-5\right)=-6$$

Use the distributive property to multiply n by -n-5.

$$-n^{2}-5n=-6$$

Divide both sides by -1.

$$\frac{-n^{2}-5n}{-1}=-\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\left(-\frac{5}{-1}\right)n=-\frac{6}{-1}$$

Divide -5 by -1.

$$n^{2}+5n=-\frac{6}{-1}$$

Divide -6 by -1.

$$n^{2}+5n=6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+5n+\left(\frac{5}{2}\right)^{2}=6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+5n+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$n^{2}+5n+\frac{25}{4}=\frac{49}{4}$$

Factor n^{2}+5n+\frac{25}{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}.

$$\left(n+\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$n+\frac{5}{2}=\frac{7}{2} n+\frac{5}{2}=-\frac{7}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$n=1 n=-6$$