Solve (2+2n)n=2019*2

$$( 2 + 2 n ) n = 2019 \times 2$$

Solve for n

$$n = \frac{\sqrt{8077} - 1}{2} \approx 44.436065693$$

$$n=\frac{-\sqrt{8077}-1}{2}\approx -45.436065693$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2+2n by n.

$$2n+2n^{2}=2019\times 2$$

Multiply 2019 and 2 to get 4038.

$$2n+2n^{2}=4038$$

Subtract 4038 from both sides.

$$2n+2n^{2}-4038=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2n^{2}+2n-4038=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -4038 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-2±\sqrt{2^{2}-4\times 2\left(-4038\right)}}{2\times 2}$$

Square 2.

$$n=\frac{-2±\sqrt{4-4\times 2\left(-4038\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-2±\sqrt{4-8\left(-4038\right)}}{2\times 2}$$

Multiply -8 times -4038.

$$n=\frac{-2±\sqrt{4+32304}}{2\times 2}$$

Add 4 to 32304.

$$n=\frac{-2±\sqrt{32308}}{2\times 2}$$

Take the square root of 32308.

$$n=\frac{-2±2\sqrt{8077}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-2±2\sqrt{8077}}{4}$$

Now solve the equation n=\frac{-2±2\sqrt{8077}}{4} when ± is plus. Add -2 to 2\sqrt{8077}.

$$n=\frac{2\sqrt{8077}-2}{4}$$

Divide -2+2\sqrt{8077} by 4.

$$n=\frac{\sqrt{8077}-1}{2}$$

Now solve the equation n=\frac{-2±2\sqrt{8077}}{4} when ± is minus. Subtract 2\sqrt{8077} from -2.

$$n=\frac{-2\sqrt{8077}-2}{4}$$

Divide -2-2\sqrt{8077} by 4.

$$n=\frac{-\sqrt{8077}-1}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{8077}-1}{2} n=\frac{-\sqrt{8077}-1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2+2n by n.

$$2n+2n^{2}=2019\times 2$$

Multiply 2019 and 2 to get 4038.

$$2n+2n^{2}=4038$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2n^{2}+2n=4038$$

Divide both sides by 2.

$$\frac{2n^{2}+2n}{2}=\frac{4038}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\frac{2}{2}n=\frac{4038}{2}$$

Divide 2 by 2.

$$n^{2}+n=\frac{4038}{2}$$

Divide 4038 by 2.

$$n^{2}+n=2019$$

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.

$$n^{2}+n+\left(\frac{1}{2}\right)^{2}=2019+\left(\frac{1}{2}\right)^{2}$$

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

$$n^{2}+n+\frac{1}{4}=2019+\frac{1}{4}$$

Add 2019 to \frac{1}{4}.

$$n^{2}+n+\frac{1}{4}=\frac{8077}{4}$$

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

$$\left(n+\frac{1}{2}\right)^{2}=\frac{8077}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{2}\right)^{2}}=\sqrt{\frac{8077}{4}}$$

Simplify.

$$n+\frac{1}{2}=\frac{\sqrt{8077}}{2} n+\frac{1}{2}=-\frac{\sqrt{8077}}{2}$$

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

$$n=\frac{\sqrt{8077}-1}{2} n=\frac{-\sqrt{8077}-1}{2}$$