Solve a^2-7a=a+20

$$a ^ { 2 } - 7 a = a + 20$$

Solve for a

$$a=-2$$

$$a=10$$

Steps Using Factoring

Subtract a from both sides.

$$a^{2}-7a-a=20$$

Combine -7a and -a to get -8a.

$$a^{2}-8a=20$$

Subtract 20 from both sides.

$$a^{2}-8a-20=0$$

To solve the equation, factor a^{2}-8a-20 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

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

$$\left(a-10\right)\left(a+2\right)$$

To find equation solutions, solve a-10=0 and a+2=0.

$$a=10 a=-2$$

Steps Using Factoring By Grouping

Subtract a from both sides.

$$a^{2}-7a-a=20$$

Combine -7a and -a to get -8a.

$$a^{2}-8a=20$$

Subtract 20 from both sides.

$$a^{2}-8a-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-20. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-20\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

Rewrite a^{2}-8a-20 as \left(a^{2}-10a\right)+\left(2a-20\right).

$$\left(a^{2}-10a\right)+\left(2a-20\right)$$

Factor out a in the first and 2 in the second group.

$$a\left(a-10\right)+2\left(a-10\right)$$

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

$$\left(a-10\right)\left(a+2\right)$$

To find equation solutions, solve a-10=0 and a+2=0.

$$a=10 a=-2$$

Steps Using the Quadratic Formula

Subtract a from both sides.

$$a^{2}-7a-a=20$$

Combine -7a and -a to get -8a.

$$a^{2}-8a=20$$

Subtract 20 from both sides.

$$a^{2}-8a-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-20\right)}}{2}$$

Square -8.

$$a=\frac{-\left(-8\right)±\sqrt{64-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$a=\frac{-\left(-8\right)±\sqrt{64+80}}{2}$$

Add 64 to 80.

$$a=\frac{-\left(-8\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$a=\frac{-\left(-8\right)±12}{2}$$

The opposite of -8 is 8.

$$a=\frac{8±12}{2}$$

Now solve the equation a=\frac{8±12}{2} when ± is plus. Add 8 to 12.

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

Divide 20 by 2.

$$a=10$$

Now solve the equation a=\frac{8±12}{2} when ± is minus. Subtract 12 from 8.

$$a=-\frac{4}{2}$$

Divide -4 by 2.

$$a=-2$$

The equation is now solved.

$$a=10 a=-2$$

Steps for Completing the Square

Subtract a from both sides.

$$a^{2}-7a-a=20$$

Combine -7a and -a to get -8a.

$$a^{2}-8a=20$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-8a+\left(-4\right)^{2}=20+\left(-4\right)^{2}$$

Square -4.

$$a^{2}-8a+16=20+16$$

Add 20 to 16.

$$a^{2}-8a+16=36$$

Factor a^{2}-8a+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-4\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-4\right)^{2}}=\sqrt{36}$$

Simplify.

$$a-4=6 a-4=-6$$

Add 4 to both sides of the equation.

$$a=10 a=-2$$