Solve (400(1+2x+x^2)+800x^2)*2=1.1^2*2800

$$(400(1+2x+ { x }^{ 2 } )+800 { x }^{ 2 } ) \times 2= { 1.1 }^{ 2 } \times 2800$$

Solve for x

$$x=\frac{\sqrt{4282}}{60}-\frac{1}{3}\approx 0.75728321$$

$$x=-\frac{\sqrt{4282}}{60}-\frac{1}{3}\approx -1.423949877$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 400 by 1+2x+x^{2}.

$$\left(400+800x+400x^{2}+800x^{2}\right)\times 2=1.1^{2}\times 2800$$

Combine 400x^{2} and 800x^{2} to get 1200x^{2}.

$$\left(400+800x+1200x^{2}\right)\times 2=1.1^{2}\times 2800$$

Use the distributive property to multiply 400+800x+1200x^{2} by 2.

$$800+1600x+2400x^{2}=1.1^{2}\times 2800$$

Calculate 1.1 to the power of 2 and get 1.21.

$$800+1600x+2400x^{2}=1.21\times 2800$$

Multiply 1.21 and 2800 to get 3388.

$$800+1600x+2400x^{2}=3388$$

Subtract 3388 from both sides.

$$800+1600x+2400x^{2}-3388=0$$

Subtract 3388 from 800 to get -2588.

$$-2588+1600x+2400x^{2}=0$$

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.

$$2400x^{2}+1600x-2588=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2400 for a, 1600 for b, and -2588 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1600±\sqrt{1600^{2}-4\times 2400\left(-2588\right)}}{2\times 2400}$$

Square 1600.

$$x=\frac{-1600±\sqrt{2560000-4\times 2400\left(-2588\right)}}{2\times 2400}$$

Multiply -4 times 2400.

$$x=\frac{-1600±\sqrt{2560000-9600\left(-2588\right)}}{2\times 2400}$$

Multiply -9600 times -2588.

$$x=\frac{-1600±\sqrt{2560000+24844800}}{2\times 2400}$$

Add 2560000 to 24844800.

$$x=\frac{-1600±\sqrt{27404800}}{2\times 2400}$$

Take the square root of 27404800.

$$x=\frac{-1600±80\sqrt{4282}}{2\times 2400}$$

Multiply 2 times 2400.

$$x=\frac{-1600±80\sqrt{4282}}{4800}$$

Now solve the equation x=\frac{-1600±80\sqrt{4282}}{4800} when ± is plus. Add -1600 to 80\sqrt{4282}.

$$x=\frac{80\sqrt{4282}-1600}{4800}$$

Divide -1600+80\sqrt{4282} by 4800.

$$x=\frac{\sqrt{4282}}{60}-\frac{1}{3}$$

Now solve the equation x=\frac{-1600±80\sqrt{4282}}{4800} when ± is minus. Subtract 80\sqrt{4282} from -1600.

$$x=\frac{-80\sqrt{4282}-1600}{4800}$$

Divide -1600-80\sqrt{4282} by 4800.

$$x=-\frac{\sqrt{4282}}{60}-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{4282}}{60}-\frac{1}{3} x=-\frac{\sqrt{4282}}{60}-\frac{1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 400 by 1+2x+x^{2}.

$$\left(400+800x+400x^{2}+800x^{2}\right)\times 2=1.1^{2}\times 2800$$

Combine 400x^{2} and 800x^{2} to get 1200x^{2}.

$$\left(400+800x+1200x^{2}\right)\times 2=1.1^{2}\times 2800$$

Use the distributive property to multiply 400+800x+1200x^{2} by 2.

$$800+1600x+2400x^{2}=1.1^{2}\times 2800$$

Calculate 1.1 to the power of 2 and get 1.21.

$$800+1600x+2400x^{2}=1.21\times 2800$$

Multiply 1.21 and 2800 to get 3388.

$$800+1600x+2400x^{2}=3388$$

Subtract 800 from both sides.

$$1600x+2400x^{2}=3388-800$$

Subtract 800 from 3388 to get 2588.

$$1600x+2400x^{2}=2588$$

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.

$$2400x^{2}+1600x=2588$$

Divide both sides by 2400.

$$\frac{2400x^{2}+1600x}{2400}=\frac{2588}{2400}$$

Dividing by 2400 undoes the multiplication by 2400.

$$x^{2}+\frac{1600}{2400}x=\frac{2588}{2400}$$

Reduce the fraction \frac{1600}{2400} to lowest terms by extracting and canceling out 800.

$$x^{2}+\frac{2}{3}x=\frac{2588}{2400}$$

Reduce the fraction \frac{2588}{2400} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{2}{3}x=\frac{647}{600}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{647}{600}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{647}{600}+\frac{1}{9}$$

Add \frac{647}{600} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{2141}{1800}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{2141}{1800}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{2141}{1800}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{4282}}{60} x+\frac{1}{3}=-\frac{\sqrt{4282}}{60}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{\sqrt{4282}}{60}-\frac{1}{3} x=-\frac{\sqrt{4282}}{60}-\frac{1}{3}$$