Solve n^2-13n+3=-27

$$n ^ { 2 } - 13 n + 3 = - 27$$

Solve for n

$$n=3$$

$$n=10$$

Steps Using Factoring

Add 27 to both sides.

$$n^{2}-13n+3+27=0$$

Add 3 and 27 to get 30.

$$n^{2}-13n+30=0$$

To solve the equation, factor n^{2}-13n+30 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.

$$a+b=-13 ab=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

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

$$\left(n-10\right)\left(n-3\right)$$

To find equation solutions, solve n-10=0 and n-3=0.

$$n=10 n=3$$

Steps Using Factoring By Grouping

Add 27 to both sides.

$$n^{2}-13n+3+27=0$$

Add 3 and 27 to get 30.

$$n^{2}-13n+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn+30. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 30=30$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -13.

$$a=-10 b=-3$$

Rewrite n^{2}-13n+30 as \left(n^{2}-10n\right)+\left(-3n+30\right).

$$\left(n^{2}-10n\right)+\left(-3n+30\right)$$

Factor out n in the first and -3 in the second group.

$$n\left(n-10\right)-3\left(n-10\right)$$

Factor out common term n-10 by using distributive property.

$$\left(n-10\right)\left(n-3\right)$$

To find equation solutions, solve n-10=0 and n-3=0.

$$n=10 n=3$$

Steps Using the Quadratic Formula

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.

$$n^{2}-13n+3=-27$$

Add 27 to both sides of the equation.

$$n^{2}-13n+3-\left(-27\right)=-27-\left(-27\right)$$

Subtracting -27 from itself leaves 0.

$$n^{2}-13n+3-\left(-27\right)=0$$

Subtract -27 from 3.

$$n^{2}-13n+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 30}}{2}$$

Square -13.

$$n=\frac{-\left(-13\right)±\sqrt{169-4\times 30}}{2}$$

Multiply -4 times 30.

$$n=\frac{-\left(-13\right)±\sqrt{169-120}}{2}$$

Add 169 to -120.

$$n=\frac{-\left(-13\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$n=\frac{-\left(-13\right)±7}{2}$$

The opposite of -13 is 13.

$$n=\frac{13±7}{2}$$

Now solve the equation n=\frac{13±7}{2} when ± is plus. Add 13 to 7.

$$n=\frac{20}{2}$$

Divide 20 by 2.

$$n=10$$

Now solve the equation n=\frac{13±7}{2} when ± is minus. Subtract 7 from 13.

$$n=\frac{6}{2}$$

Divide 6 by 2.

$$n=3$$

The equation is now solved.

$$n=10 n=3$$

Steps for Completing the Square

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.

$$n^{2}-13n+3=-27$$

Subtract 3 from both sides of the equation.

$$n^{2}-13n+3-3=-27-3$$

Subtracting 3 from itself leaves 0.

$$n^{2}-13n=-27-3$$

Subtract 3 from -27.

$$n^{2}-13n=-30$$

Divide -13, the coefficient of the x term, by 2 to get -\frac{13}{2}. Then add the square of -\frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-13n+\left(-\frac{13}{2}\right)^{2}=-30+\left(-\frac{13}{2}\right)^{2}$$

Square -\frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-13n+\frac{169}{4}=-30+\frac{169}{4}$$

Add -30 to \frac{169}{4}.

$$n^{2}-13n+\frac{169}{4}=\frac{49}{4}$$

Factor n^{2}-13n+\frac{169}{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}.

$$\left(n-\frac{13}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{13}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$n-\frac{13}{2}=\frac{7}{2} n-\frac{13}{2}=-\frac{7}{2}$$

Add \frac{13}{2} to both sides of the equation.

$$n=10 n=3$$