Solve 0=b^2-13b+42

$$0 = b ^ { 2 } - 13 b + 42$$

Solve for b

$$b=6$$

$$b=7$$

Steps Using Factoring

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

$$b^{2}-13b+42=0$$

To solve the equation, factor b^{2}-13b+42 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.

$$a+b=-13 ab=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

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

$$\left(b-7\right)\left(b-6\right)$$

To find equation solutions, solve b-7=0 and b-6=0.

$$b=7 b=6$$

Steps Using Factoring By Grouping

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

$$b^{2}-13b+42=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb+42. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 42=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

Rewrite b^{2}-13b+42 as \left(b^{2}-7b\right)+\left(-6b+42\right).

$$\left(b^{2}-7b\right)+\left(-6b+42\right)$$

Factor out b in the first and -6 in the second group.

$$b\left(b-7\right)-6\left(b-7\right)$$

Factor out common term b-7 by using distributive property.

$$\left(b-7\right)\left(b-6\right)$$

To find equation solutions, solve b-7=0 and b-6=0.

$$b=7 b=6$$

Steps Using the Quadratic Formula

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

$$b^{2}-13b+42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -13 for b, and 42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 42}}{2}$$

Square -13.

$$b=\frac{-\left(-13\right)±\sqrt{169-4\times 42}}{2}$$

Multiply -4 times 42.

$$b=\frac{-\left(-13\right)±\sqrt{169-168}}{2}$$

Add 169 to -168.

$$b=\frac{-\left(-13\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$b=\frac{-\left(-13\right)±1}{2}$$

The opposite of -13 is 13.

$$b=\frac{13±1}{2}$$

Now solve the equation b=\frac{13±1}{2} when ± is plus. Add 13 to 1.

$$b=\frac{14}{2}$$

Divide 14 by 2.

$$b=7$$

Now solve the equation b=\frac{13±1}{2} when ± is minus. Subtract 1 from 13.

$$b=\frac{12}{2}$$

Divide 12 by 2.

$$b=6$$

The equation is now solved.

$$b=7 b=6$$

Steps for Completing the Square

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

$$b^{2}-13b+42=0$$

Subtract 42 from both sides. Anything subtracted from zero gives its negation.

$$b^{2}-13b=-42$$

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.

$$b^{2}-13b+\left(-\frac{13}{2}\right)^{2}=-42+\left(-\frac{13}{2}\right)^{2}$$

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

$$b^{2}-13b+\frac{169}{4}=-42+\frac{169}{4}$$

Add -42 to \frac{169}{4}.

$$b^{2}-13b+\frac{169}{4}=\frac{1}{4}$$

Factor b^{2}-13b+\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}.

$$\left(b-\frac{13}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$b-\frac{13}{2}=\frac{1}{2} b-\frac{13}{2}=-\frac{1}{2}$$

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

$$b=7 b=6$$