Solve (2a-2)^2+a^2-a-2=16

$$( 2 a - 2 ) ^ { 2 } + a ^ { 2 } - a - 2 = 16$$

Solve for a

$$a=-1$$

$$a = \frac{14}{5} = 2\frac{4}{5} = 2.8$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-2\right)^{2}.

$$4a^{2}-8a+4+a^{2}-a-2=16$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}-8a+4-a-2=16$$

Combine -8a and -a to get -9a.

$$5a^{2}-9a+4-2=16$$

Subtract 2 from 4 to get 2.

$$5a^{2}-9a+2=16$$

Subtract 16 from both sides.

$$5a^{2}-9a+2-16=0$$

Subtract 16 from 2 to get -14.

$$5a^{2}-9a-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5a^{2}+aa+ba-14. To find a and b, set up a system to be solved.

$$a+b=-9 ab=5\left(-14\right)=-70$$

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 -70.

$$1,-70 2,-35 5,-14 7,-10$$

Calculate the sum for each pair.

$$1-70=-69 2-35=-33 5-14=-9 7-10=-3$$

The solution is the pair that gives sum -9.

$$a=-14 b=5$$

Rewrite 5a^{2}-9a-14 as \left(5a^{2}-14a\right)+\left(5a-14\right).

$$\left(5a^{2}-14a\right)+\left(5a-14\right)$$

Factor out a in 5a^{2}-14a.

$$a\left(5a-14\right)+5a-14$$

Factor out common term 5a-14 by using distributive property.

$$\left(5a-14\right)\left(a+1\right)$$

To find equation solutions, solve 5a-14=0 and a+1=0.

$$a=\frac{14}{5} a=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-2\right)^{2}.

$$4a^{2}-8a+4+a^{2}-a-2=16$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}-8a+4-a-2=16$$

Combine -8a and -a to get -9a.

$$5a^{2}-9a+4-2=16$$

Subtract 2 from 4 to get 2.

$$5a^{2}-9a+2=16$$

Subtract 16 from both sides.

$$5a^{2}-9a+2-16=0$$

Subtract 16 from 2 to get -14.

$$5a^{2}-9a-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -9 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-14\right)}}{2\times 5}$$

Square -9.

$$a=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-14\right)}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-\left(-9\right)±\sqrt{81-20\left(-14\right)}}{2\times 5}$$

Multiply -20 times -14.

$$a=\frac{-\left(-9\right)±\sqrt{81+280}}{2\times 5}$$

Add 81 to 280.

$$a=\frac{-\left(-9\right)±\sqrt{361}}{2\times 5}$$

Take the square root of 361.

$$a=\frac{-\left(-9\right)±19}{2\times 5}$$

The opposite of -9 is 9.

$$a=\frac{9±19}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{9±19}{10}$$

Now solve the equation a=\frac{9±19}{10} when ± is plus. Add 9 to 19.

$$a=\frac{28}{10}$$

Reduce the fraction \frac{28}{10} to lowest terms by extracting and canceling out 2.

$$a=\frac{14}{5}$$

Now solve the equation a=\frac{9±19}{10} when ± is minus. Subtract 19 from 9.

$$a=-\frac{10}{10}$$

Divide -10 by 10.

$$a=-1$$

The equation is now solved.

$$a=\frac{14}{5} a=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2a-2\right)^{2}.

$$4a^{2}-8a+4+a^{2}-a-2=16$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$5a^{2}-8a+4-a-2=16$$

Combine -8a and -a to get -9a.

$$5a^{2}-9a+4-2=16$$

Subtract 2 from 4 to get 2.

$$5a^{2}-9a+2=16$$

Subtract 2 from both sides.

$$5a^{2}-9a=16-2$$

Subtract 2 from 16 to get 14.

$$5a^{2}-9a=14$$

Divide both sides by 5.

$$\frac{5a^{2}-9a}{5}=\frac{14}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$a^{2}-\frac{9}{5}a=\frac{14}{5}$$

Divide -\frac{9}{5}, the coefficient of the x term, by 2 to get -\frac{9}{10}. Then add the square of -\frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{9}{5}a+\left(-\frac{9}{10}\right)^{2}=\frac{14}{5}+\left(-\frac{9}{10}\right)^{2}$$

Square -\frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{9}{5}a+\frac{81}{100}=\frac{14}{5}+\frac{81}{100}$$

Add \frac{14}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{9}{5}a+\frac{81}{100}=\frac{361}{100}$$

Factor a^{2}-\frac{9}{5}a+\frac{81}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{9}{10}\right)^{2}=\frac{361}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{9}{10}\right)^{2}}=\sqrt{\frac{361}{100}}$$

Simplify.

$$a-\frac{9}{10}=\frac{19}{10} a-\frac{9}{10}=-\frac{19}{10}$$

Add \frac{9}{10} to both sides of the equation.

$$a=\frac{14}{5} a=-1$$