Solve 1+3=n-n^2-7n/2

$$1+3 = n- \frac{ { n }^{ 2 } -7n }{ 2 }$$

Solve for n

$$n=1$$

$$n=8$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

$$2+6=2n-\left(n^{2}-7n\right)$$

Add 2 and 6 to get 8.

$$8=2n-\left(n^{2}-7n\right)$$

To find the opposite of n^{2}-7n, find the opposite of each term.

$$8=2n-n^{2}+7n$$

Combine 2n and 7n to get 9n.

$$8=9n-n^{2}$$

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

$$9n-n^{2}=8$$

Subtract 8 from both sides.

$$9n-n^{2}-8=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-n^{2}+9n-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-8. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-8\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 9.

$$a=8 b=1$$

Rewrite -n^{2}+9n-8 as \left(-n^{2}+8n\right)+\left(n-8\right).

$$\left(-n^{2}+8n\right)+\left(n-8\right)$$

Factor out -n in -n^{2}+8n.

$$-n\left(n-8\right)+n-8$$

Factor out common term n-8 by using distributive property.

$$\left(n-8\right)\left(-n+1\right)$$

To find equation solutions, solve n-8=0 and -n+1=0.

$$n=8 n=1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2+6=2n-\left(n^{2}-7n\right)$$

Add 2 and 6 to get 8.

$$8=2n-\left(n^{2}-7n\right)$$

To find the opposite of n^{2}-7n, find the opposite of each term.

$$8=2n-n^{2}+7n$$

Combine 2n and 7n to get 9n.

$$8=9n-n^{2}$$

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

$$9n-n^{2}=8$$

Subtract 8 from both sides.

$$9n-n^{2}-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-n^{2}+9n-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Square 9.

$$n=\frac{-9±\sqrt{81-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-9±\sqrt{81+4\left(-8\right)}}{2\left(-1\right)}$$

Multiply 4 times -8.

$$n=\frac{-9±\sqrt{81-32}}{2\left(-1\right)}$$

Add 81 to -32.

$$n=\frac{-9±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$n=\frac{-9±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-9±7}{-2}$$

Now solve the equation n=\frac{-9±7}{-2} when ± is plus. Add -9 to 7.

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

Divide -2 by -2.

$$n=1$$

Now solve the equation n=\frac{-9±7}{-2} when ± is minus. Subtract 7 from -9.

$$n=-\frac{16}{-2}$$

Divide -16 by -2.

$$n=8$$

The equation is now solved.

$$n=1 n=8$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2+6=2n-\left(n^{2}-7n\right)$$

Add 2 and 6 to get 8.

$$8=2n-\left(n^{2}-7n\right)$$

To find the opposite of n^{2}-7n, find the opposite of each term.

$$8=2n-n^{2}+7n$$

Combine 2n and 7n to get 9n.

$$8=9n-n^{2}$$

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

$$9n-n^{2}=8$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-n^{2}+9n=8$$

Divide both sides by -1.

$$\frac{-n^{2}+9n}{-1}=\frac{8}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{9}{-1}n=\frac{8}{-1}$$

Divide 9 by -1.

$$n^{2}-9n=\frac{8}{-1}$$

Divide 8 by -1.

$$n^{2}-9n=-8$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-9n+\left(-\frac{9}{2}\right)^{2}=-8+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-9n+\frac{81}{4}=-8+\frac{81}{4}$$

Add -8 to \frac{81}{4}.

$$n^{2}-9n+\frac{81}{4}=\frac{49}{4}$$

Factor n^{2}-9n+\frac{81}{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}.

$$\left(n-\frac{9}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{9}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$n-\frac{9}{2}=\frac{7}{2} n-\frac{9}{2}=-\frac{7}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$n=8 n=1$$