Solve 2a(a-3)=a-6

$$2 a ( a - 3 ) = a - 6$$

Solve for a

$$a=2$$

$$a = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2a by a-3.

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

Subtract a from both sides.

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

Combine -6a and -a to get -7a.

$$2a^{2}-7a=-6$$

Add 6 to both sides.

$$2a^{2}-7a+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Square -7.

$$a=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-7\right)±\sqrt{49-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$a=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 2}$$

Add 49 to -48.

$$a=\frac{-\left(-7\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$a=\frac{-\left(-7\right)±1}{2\times 2}$$

The opposite of -7 is 7.

$$a=\frac{7±1}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{7±1}{4}$$

Now solve the equation a=\frac{7±1}{4} when ± is plus. Add 7 to 1.

$$a=\frac{8}{4}$$

Divide 8 by 4.

$$a=2$$

Now solve the equation a=\frac{7±1}{4} when ± is minus. Subtract 1 from 7.

$$a=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$a=\frac{3}{2}$$

The equation is now solved.

$$a=2 a=\frac{3}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2a by a-3.

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

Subtract a from both sides.

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

Combine -6a and -a to get -7a.

$$2a^{2}-7a=-6$$

Divide both sides by 2.

$$\frac{2a^{2}-7a}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}-\frac{7}{2}a=-\frac{6}{2}$$

Divide -6 by 2.

$$a^{2}-\frac{7}{2}a=-3$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{7}{2}a+\left(-\frac{7}{4}\right)^{2}=-3+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{7}{2}a+\frac{49}{16}=-3+\frac{49}{16}$$

Add -3 to \frac{49}{16}.

$$a^{2}-\frac{7}{2}a+\frac{49}{16}=\frac{1}{16}$$

Factor a^{2}-\frac{7}{2}a+\frac{49}{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}.

$$\left(a-\frac{7}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{7}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$a-\frac{7}{4}=\frac{1}{4} a-\frac{7}{4}=-\frac{1}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$a=2 a=\frac{3}{2}$$