Solve 70=n^2-5n+4

$$70 = { n }^{ 2 } -5n+4$$

Solve for n

$$n=-6$$

$$n=11$$

Steps Using Factoring

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

$$n^{2}-5n+4=70$$

Subtract 70 from both sides.

$$n^{2}-5n+4-70=0$$

Subtract 70 from 4 to get -66.

$$n^{2}-5n-66=0$$

To solve the equation, factor n^{2}-5n-66 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=-66$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -66.

$$1,-66 2,-33 3,-22 6,-11$$

Calculate the sum for each pair.

$$1-66=-65 2-33=-31 3-22=-19 6-11=-5$$

The solution is the pair that gives sum -5.

$$a=-11 b=6$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n-11\right)\left(n+6\right)$$

To find equation solutions, solve n-11=0 and n+6=0.

$$n=11 n=-6$$

Steps Using Factoring By Grouping

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

$$n^{2}-5n+4=70$$

Subtract 70 from both sides.

$$n^{2}-5n+4-70=0$$

Subtract 70 from 4 to get -66.

$$n^{2}-5n-66=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-66. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-66\right)=-66$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -66.

$$1,-66 2,-33 3,-22 6,-11$$

Calculate the sum for each pair.

$$1-66=-65 2-33=-31 3-22=-19 6-11=-5$$

The solution is the pair that gives sum -5.

$$a=-11 b=6$$

Rewrite n^{2}-5n-66 as \left(n^{2}-11n\right)+\left(6n-66\right).

$$\left(n^{2}-11n\right)+\left(6n-66\right)$$

Factor out n in the first and 6 in the second group.

$$n\left(n-11\right)+6\left(n-11\right)$$

Factor out common term n-11 by using distributive property.

$$\left(n-11\right)\left(n+6\right)$$

To find equation solutions, solve n-11=0 and n+6=0.

$$n=11 n=-6$$

Steps Using the Quadratic Formula

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

$$n^{2}-5n+4=70$$

Subtract 70 from both sides.

$$n^{2}-5n+4-70=0$$

Subtract 70 from 4 to get -66.

$$n^{2}-5n-66=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -66 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-66\right)}}{2}$$

Square -5.

$$n=\frac{-\left(-5\right)±\sqrt{25-4\left(-66\right)}}{2}$$

Multiply -4 times -66.

$$n=\frac{-\left(-5\right)±\sqrt{25+264}}{2}$$

Add 25 to 264.

$$n=\frac{-\left(-5\right)±\sqrt{289}}{2}$$

Take the square root of 289.

$$n=\frac{-\left(-5\right)±17}{2}$$

The opposite of -5 is 5.

$$n=\frac{5±17}{2}$$

Now solve the equation n=\frac{5±17}{2} when ± is plus. Add 5 to 17.

$$n=\frac{22}{2}$$

Divide 22 by 2.

$$n=11$$

Now solve the equation n=\frac{5±17}{2} when ± is minus. Subtract 17 from 5.

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

Divide -12 by 2.

$$n=-6$$

The equation is now solved.

$$n=11 n=-6$$

Steps for Completing the Square

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

$$n^{2}-5n+4=70$$

Subtract 4 from both sides.

$$n^{2}-5n=70-4$$

Subtract 4 from 70 to get 66.

$$n^{2}-5n=66$$

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.

$$n^{2}-5n+\left(-\frac{5}{2}\right)^{2}=66+\left(-\frac{5}{2}\right)^{2}$$

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

$$n^{2}-5n+\frac{25}{4}=66+\frac{25}{4}$$

Add 66 to \frac{25}{4}.

$$n^{2}-5n+\frac{25}{4}=\frac{289}{4}$$

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}.

$$\left(n-\frac{5}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{5}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$n-\frac{5}{2}=\frac{17}{2} n-\frac{5}{2}=-\frac{17}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$n=11 n=-6$$