Solve (-2a)^2-4*1*3a=0

$$( - 2 a ) ^ { 2 } - 4 \cdot 1 \cdot 3 a = 0$$

Solve for a

$$a=3$$

$$a=0$$

Steps Using Factoring

Expand \left(-2a\right)^{2}.

$$\left(-2\right)^{2}a^{2}-4\times 1\times 3a=0$$

Calculate -2 to the power of 2 and get 4.

$$4a^{2}-4\times 1\times 3a=0$$

Multiply 4 and 1 to get 4.

$$4a^{2}-4\times 3a=0$$

Multiply 4 and 3 to get 12.

$$4a^{2}-12a=0$$

Factor out a.

$$a\left(4a-12\right)=0$$

To find equation solutions, solve a=0 and 4a-12=0.

$$a=0 a=3$$

Steps Using the Quadratic Formula

Expand \left(-2a\right)^{2}.

$$\left(-2\right)^{2}a^{2}-4\times 1\times 3a=0$$

Calculate -2 to the power of 2 and get 4.

$$4a^{2}-4\times 1\times 3a=0$$

Multiply 4 and 1 to get 4.

$$4a^{2}-4\times 3a=0$$

Multiply 4 and 3 to get 12.

$$4a^{2}-12a=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 4}$$

Take the square root of \left(-12\right)^{2}.

$$a=\frac{-\left(-12\right)±12}{2\times 4}$$

The opposite of -12 is 12.

$$a=\frac{12±12}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{12±12}{8}$$

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

$$a=\frac{24}{8}$$

Divide 24 by 8.

$$a=3$$

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

$$a=\frac{0}{8}$$

Divide 0 by 8.

$$a=0$$

The equation is now solved.

$$a=3 a=0$$

Steps for Completing the Square

Expand \left(-2a\right)^{2}.

$$\left(-2\right)^{2}a^{2}-4\times 1\times 3a=0$$

Calculate -2 to the power of 2 and get 4.

$$4a^{2}-4\times 1\times 3a=0$$

Multiply 4 and 1 to get 4.

$$4a^{2}-4\times 3a=0$$

Multiply 4 and 3 to get 12.

$$4a^{2}-12a=0$$

Divide both sides by 4.

$$\frac{4a^{2}-12a}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$a^{2}+\left(-\frac{12}{4}\right)a=\frac{0}{4}$$

Divide -12 by 4.

$$a^{2}-3a=\frac{0}{4}$$

Divide 0 by 4.

$$a^{2}-3a=0$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-3a+\frac{9}{4}=\frac{9}{4}$$

Factor a^{2}-3a+\frac{9}{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}.

$$\left(a-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$a=3 a=0$$