Solve (n-3)^2+4n^2=4

$$( n - 3 ) ^ { 2 } + 4 n ^ { 2 } = 4$$

Solve for n

$$n=\frac{3}{5}+\frac{4}{5}i=0.6+0.8i$$

$$n=\frac{3}{5}-\frac{4}{5}i=0.6-0.8i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-3\right)^{2}.

$$n^{2}-6n+9+4n^{2}=4$$

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

$$5n^{2}-6n+9=4$$

Subtract 4 from both sides.

$$5n^{2}-6n+9-4=0$$

Subtract 4 from 9 to get 5.

$$5n^{2}-6n+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Square -6.

$$n=\frac{-\left(-6\right)±\sqrt{36-4\times 5\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-\left(-6\right)±\sqrt{36-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$n=\frac{-\left(-6\right)±\sqrt{36-100}}{2\times 5}$$

Add 36 to -100.

$$n=\frac{-\left(-6\right)±\sqrt{-64}}{2\times 5}$$

Take the square root of -64.

$$n=\frac{-\left(-6\right)±8i}{2\times 5}$$

The opposite of -6 is 6.

$$n=\frac{6±8i}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{6±8i}{10}$$

Now solve the equation n=\frac{6±8i}{10} when ± is plus. Add 6 to 8i.

$$n=\frac{6+8i}{10}$$

Divide 6+8i by 10.

$$n=\frac{3}{5}+\frac{4}{5}i$$

Now solve the equation n=\frac{6±8i}{10} when ± is minus. Subtract 8i from 6.

$$n=\frac{6-8i}{10}$$

Divide 6-8i by 10.

$$n=\frac{3}{5}-\frac{4}{5}i$$

The equation is now solved.

$$n=\frac{3}{5}+\frac{4}{5}i n=\frac{3}{5}-\frac{4}{5}i$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-3\right)^{2}.

$$n^{2}-6n+9+4n^{2}=4$$

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

$$5n^{2}-6n+9=4$$

Subtract 9 from both sides.

$$5n^{2}-6n=4-9$$

Subtract 9 from 4 to get -5.

$$5n^{2}-6n=-5$$

Divide both sides by 5.

$$\frac{5n^{2}-6n}{5}=-\frac{5}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$n^{2}-\frac{6}{5}n=-\frac{5}{5}$$

Divide -5 by 5.

$$n^{2}-\frac{6}{5}n=-1$$

Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{6}{5}n+\left(-\frac{3}{5}\right)^{2}=-1+\left(-\frac{3}{5}\right)^{2}$$

Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{6}{5}n+\frac{9}{25}=-1+\frac{9}{25}$$

Add -1 to \frac{9}{25}.

$$n^{2}-\frac{6}{5}n+\frac{9}{25}=-\frac{16}{25}$$

Factor n^{2}-\frac{6}{5}n+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{3}{5}\right)^{2}=-\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{5}\right)^{2}}=\sqrt{-\frac{16}{25}}$$

Simplify.

$$n-\frac{3}{5}=\frac{4}{5}i n-\frac{3}{5}=-\frac{4}{5}i$$

Add \frac{3}{5} to both sides of the equation.

$$n=\frac{3}{5}+\frac{4}{5}i n=\frac{3}{5}-\frac{4}{5}i$$