Solve 20(a+3)(a-2)=a^2+a-6

$$20 ( a + 3 ) ( a - 2 ) = a ^ { 2 } + a - 6$$

Solve for a

$$a=-3$$

$$a=2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 20 by a+3.

$$\left(20a+60\right)\left(a-2\right)=a^{2}+a-6$$

Use the distributive property to multiply 20a+60 by a-2 and combine like terms.

$$20a^{2}+20a-120=a^{2}+a-6$$

Subtract a^{2} from both sides.

$$20a^{2}+20a-120-a^{2}=a-6$$

Combine 20a^{2} and -a^{2} to get 19a^{2}.

$$19a^{2}+20a-120=a-6$$

Subtract a from both sides.

$$19a^{2}+20a-120-a=-6$$

Combine 20a and -a to get 19a.

$$19a^{2}+19a-120=-6$$

Add 6 to both sides.

$$19a^{2}+19a-120+6=0$$

Add -120 and 6 to get -114.

$$19a^{2}+19a-114=0$$

Divide both sides by 19.

$$a^{2}+a-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite a^{2}+a-6 as \left(a^{2}-2a\right)+\left(3a-6\right).

$$\left(a^{2}-2a\right)+\left(3a-6\right)$$

Factor out a in the first and 3 in the second group.

$$a\left(a-2\right)+3\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(a+3\right)$$

To find equation solutions, solve a-2=0 and a+3=0.

$$a=2 a=-3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20 by a+3.

$$\left(20a+60\right)\left(a-2\right)=a^{2}+a-6$$

Use the distributive property to multiply 20a+60 by a-2 and combine like terms.

$$20a^{2}+20a-120=a^{2}+a-6$$

Subtract a^{2} from both sides.

$$20a^{2}+20a-120-a^{2}=a-6$$

Combine 20a^{2} and -a^{2} to get 19a^{2}.

$$19a^{2}+20a-120=a-6$$

Subtract a from both sides.

$$19a^{2}+20a-120-a=-6$$

Combine 20a and -a to get 19a.

$$19a^{2}+19a-120=-6$$

Add 6 to both sides.

$$19a^{2}+19a-120+6=0$$

Add -120 and 6 to get -114.

$$19a^{2}+19a-114=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 19 for a, 19 for b, and -114 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-19±\sqrt{19^{2}-4\times 19\left(-114\right)}}{2\times 19}$$

Square 19.

$$a=\frac{-19±\sqrt{361-4\times 19\left(-114\right)}}{2\times 19}$$

Multiply -4 times 19.

$$a=\frac{-19±\sqrt{361-76\left(-114\right)}}{2\times 19}$$

Multiply -76 times -114.

$$a=\frac{-19±\sqrt{361+8664}}{2\times 19}$$

Add 361 to 8664.

$$a=\frac{-19±\sqrt{9025}}{2\times 19}$$

Take the square root of 9025.

$$a=\frac{-19±95}{2\times 19}$$

Multiply 2 times 19.

$$a=\frac{-19±95}{38}$$

Now solve the equation a=\frac{-19±95}{38} when ± is plus. Add -19 to 95.

$$a=\frac{76}{38}$$

Divide 76 by 38.

$$a=2$$

Now solve the equation a=\frac{-19±95}{38} when ± is minus. Subtract 95 from -19.

$$a=-\frac{114}{38}$$

Divide -114 by 38.

$$a=-3$$

The equation is now solved.

$$a=2 a=-3$$

Steps for Completing the Square

Use the distributive property to multiply 20 by a+3.

$$\left(20a+60\right)\left(a-2\right)=a^{2}+a-6$$

Use the distributive property to multiply 20a+60 by a-2 and combine like terms.

$$20a^{2}+20a-120=a^{2}+a-6$$

Subtract a^{2} from both sides.

$$20a^{2}+20a-120-a^{2}=a-6$$

Combine 20a^{2} and -a^{2} to get 19a^{2}.

$$19a^{2}+20a-120=a-6$$

Subtract a from both sides.

$$19a^{2}+20a-120-a=-6$$

Combine 20a and -a to get 19a.

$$19a^{2}+19a-120=-6$$

Add 120 to both sides.

$$19a^{2}+19a=-6+120$$

Add -6 and 120 to get 114.

$$19a^{2}+19a=114$$

Divide both sides by 19.

$$\frac{19a^{2}+19a}{19}=\frac{114}{19}$$

Dividing by 19 undoes the multiplication by 19.

$$a^{2}+\frac{19}{19}a=\frac{114}{19}$$

Divide 19 by 19.

$$a^{2}+a=\frac{114}{19}$$

Divide 114 by 19.

$$a^{2}+a=6$$

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.

$$a^{2}+a+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

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

$$a^{2}+a+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$a^{2}+a+\frac{1}{4}=\frac{25}{4}$$

Factor a^{2}+a+\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}.

$$\left(a+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$a+\frac{1}{2}=\frac{5}{2} a+\frac{1}{2}=-\frac{5}{2}$$

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

$$a=2 a=-3$$