Solve 42=x^2+x

$$42 = x ^ { 2 } + x$$

Solve for x

$$x=-7$$

$$x=6$$

Steps Using Factoring

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

$$x^{2}+x=42$$

Subtract 42 from both sides.

$$x^{2}+x-42=0$$

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

$$a+b=1 ab=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -42.

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

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

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

$$\left(x-6\right)\left(x+7\right)$$

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

$$x=6 x=-7$$

Steps Using Factoring By Grouping

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

$$x^{2}+x=42$$

Subtract 42 from both sides.

$$x^{2}+x-42=0$$

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

$$a+b=1 ab=1\left(-42\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -42.

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

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

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

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

Factor out x in the first and 7 in the second group.

$$x\left(x-6\right)+7\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+7\right)$$

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

$$x=6 x=-7$$

Steps Using the Quadratic Formula

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

$$x^{2}+x=42$$

Subtract 42 from both sides.

$$x^{2}+x-42=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4\left(-42\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-42\right)}}{2}$$

Multiply -4 times -42.

$$x=\frac{-1±\sqrt{1+168}}{2}$$

Add 1 to 168.

$$x=\frac{-1±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-1±13}{2}$$

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

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

Divide 12 by 2.

$$x=6$$

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

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=6 x=-7$$

Steps for Completing the Square

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

$$x^{2}+x=42$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=42+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=42+\frac{1}{4}$$

Add 42 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{169}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{13}{2} x+\frac{1}{2}=-\frac{13}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=6 x=-7$$