Solve 1250*0.8(1+x)^2=1440

$$1250 \times 0.8 ( 1 + x ) ^ { 2 } = 1440$$

Solve for x

$$x=0.2$$

$$x=-2.2$$

Steps Using Factoring By Grouping

Multiply 1250 and 0.8 to get 1000.

$$1000\left(1+x\right)^{2}=1440$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1000\left(1+2x+x^{2}\right)=1440$$

Use the distributive property to multiply 1000 by 1+2x+x^{2}.

$$1000+2000x+1000x^{2}=1440$$

Subtract 1440 from both sides.

$$1000+2000x+1000x^{2}-1440=0$$

Subtract 1440 from 1000 to get -440.

$$-440+2000x+1000x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$1000x^{2}+2000x-440=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 1000x^{2}+ax+bx-440. To find a and b, set up a system to be solved.

$$a+b=2000 ab=1000\left(-440\right)=-440000$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -440000.

$$-1,440000 -2,220000 -4,110000 -5,88000 -8,55000 -10,44000 -11,40000 -16,27500 -20,22000 -22,20000 -25,17600 -32,13750 -40,11000 -44,10000 -50,8800 -55,8000 -64,6875 -80,5500 -88,5000 -100,4400 -110,4000 -125,3520 -160,2750 -176,2500 -200,2200 -220,2000 -250,1760 -275,1600 -320,1375 -352,1250 -400,1100 -440,1000 -500,880 -550,800 -625,704$$

Calculate the sum for each pair.

$$-1+440000=439999 -2+220000=219998 -4+110000=109996 -5+88000=87995 -8+55000=54992 -10+44000=43990 -11+40000=39989 -16+27500=27484 -20+22000=21980 -22+20000=19978 -25+17600=17575 -32+13750=13718 -40+11000=10960 -44+10000=9956 -50+8800=8750 -55+8000=7945 -64+6875=6811 -80+5500=5420 -88+5000=4912 -100+4400=4300 -110+4000=3890 -125+3520=3395 -160+2750=2590 -176+2500=2324 -200+2200=2000 -220+2000=1780 -250+1760=1510 -275+1600=1325 -320+1375=1055 -352+1250=898 -400+1100=700 -440+1000=560 -500+880=380 -550+800=250 -625+704=79$$

The solution is the pair that gives sum 50.

$$a=-5 b=55$$

Rewrite 1000x^{2}+2000x-440 as \left(1000x^{2}-5x\right)+\left(55x-440\right).

$$\left(1000x^{2}-5x\right)+\left(55x-440\right)$$

Factor out 5x in the first and 11 in the second group.

$$5x\left(5x-1\right)+11\left(5x-1\right)$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(5x+11\right)$$

To find equation solutions, solve 5x-1=0 and 5x+11=0.

$$x=\frac{1}{5} x=-\frac{11}{5}$$

Steps Using the Quadratic Formula

Multiply 1250 and 0.8 to get 1000.

$$1000\left(1+x\right)^{2}=1440$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1000\left(1+2x+x^{2}\right)=1440$$

Use the distributive property to multiply 1000 by 1+2x+x^{2}.

$$1000+2000x+1000x^{2}=1440$$

Subtract 1440 from both sides.

$$1000+2000x+1000x^{2}-1440=0$$

Subtract 1440 from 1000 to get -440.

$$-440+2000x+1000x^{2}=0$$

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.

$$1000x^{2}+2000x-440=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1000 for a, 2000 for b, and -440 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2000±\sqrt{2000^{2}-4\times 1000\left(-440\right)}}{2\times 1000}$$

Square 2000.

$$x=\frac{-2000±\sqrt{4000000-4\times 1000\left(-440\right)}}{2\times 1000}$$

Multiply -4 times 1000.

$$x=\frac{-2000±\sqrt{4000000-4000\left(-440\right)}}{2\times 1000}$$

Multiply -4000 times -440.

$$x=\frac{-2000±\sqrt{4000000+1760000}}{2\times 1000}$$

Add 4000000 to 1760000.

$$x=\frac{-2000±\sqrt{5760000}}{2\times 1000}$$

Take the square root of 5760000.

$$x=\frac{-2000±2400}{2\times 1000}$$

Multiply 2 times 1000.

$$x=\frac{-2000±2400}{2000}$$

Now solve the equation x=\frac{-2000±2400}{2000} when ± is plus. Add -2000 to 2400.

$$x=\frac{400}{2000}$$

Reduce the fraction \frac{400}{2000} to lowest terms by extracting and canceling out 400.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-2000±2400}{2000} when ± is minus. Subtract 2400 from -2000.

$$x=-\frac{4400}{2000}$$

Reduce the fraction \frac{-4400}{2000} to lowest terms by extracting and canceling out 400.

$$x=-\frac{11}{5}$$

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{11}{5}$$

Steps for Completing the Square

Multiply 1250 and 0.8 to get 1000.

$$1000\left(1+x\right)^{2}=1440$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$1000\left(1+2x+x^{2}\right)=1440$$

Use the distributive property to multiply 1000 by 1+2x+x^{2}.

$$1000+2000x+1000x^{2}=1440$$

Subtract 1000 from both sides.

$$2000x+1000x^{2}=1440-1000$$

Subtract 1000 from 1440 to get 440.

$$2000x+1000x^{2}=440$$

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.

$$1000x^{2}+2000x=440$$

Divide both sides by 1000.

$$\frac{1000x^{2}+2000x}{1000}=\frac{440}{1000}$$

Dividing by 1000 undoes the multiplication by 1000.

$$x^{2}+\frac{2000}{1000}x=\frac{440}{1000}$$

Divide 2000 by 1000.

$$x^{2}+2x=\frac{440}{1000}$$

Reduce the fraction \frac{440}{1000} to lowest terms by extracting and canceling out 40.

$$x^{2}+2x=\frac{11}{25}$$

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

$$x^{2}+2x+1^{2}=\frac{11}{25}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{11}{25}+1$$

Add \frac{11}{25} to 1.

$$x^{2}+2x+1=\frac{36}{25}$$

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

$$\left(x+1\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$x+1=\frac{6}{5} x+1=-\frac{6}{5}$$

Subtract 1 from both sides of the equation.

$$x=\frac{1}{5} x=-\frac{11}{5}$$