Solve 5=1/2250x^2+1/250left(x+0.2right)^2

$$5 = \frac{ 1 }{ 2 } 250 { x }^{ 2 } + \frac{ 1 }{ 2 } 50 { \left(x+0.2 \right) }^{ 2 }$$

Solve for x

$$x=\frac{2}{15}\approx 0.133333333$$

$$x=-0.2$$

Steps Using Factoring By Grouping

Multiply \frac{1}{2} and 250 to get 125.

$$5=125x^{2}+\frac{1}{2}\times 50\left(x+0.2\right)^{2}$$

Multiply \frac{1}{2} and 50 to get 25.

$$5=125x^{2}+25\left(x+0.2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+0.2\right)^{2}.

$$5=125x^{2}+25\left(x^{2}+0.4x+0.04\right)$$

Use the distributive property to multiply 25 by x^{2}+0.4x+0.04.

$$5=125x^{2}+25x^{2}+10x+1$$

Combine 125x^{2} and 25x^{2} to get 150x^{2}.

$$5=150x^{2}+10x+1$$

Swap sides so that all variable terms are on the left hand side.

$$150x^{2}+10x+1=5$$

Subtract 5 from both sides.

$$150x^{2}+10x+1-5=0$$

Subtract 5 from 1 to get -4.

$$150x^{2}+10x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 150x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=10 ab=150\left(-4\right)=-600$$

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 -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 5.

$$a=-10 b=15$$

Rewrite 150x^{2}+10x-4 as \left(150x^{2}-10x\right)+\left(15x-4\right).

$$\left(150x^{2}-10x\right)+\left(15x-4\right)$$

Factor out 5x in 150x^{2}-10x.

$$5x\left(15x-2\right)+15x-2$$

Factor out common term 15x-2 by using distributive property.

$$\left(15x-2\right)\left(5x+1\right)$$

To find equation solutions, solve 15x-2=0 and 5x+1=0.

$$x=\frac{2}{15} x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 250 to get 125.

$$5=125x^{2}+\frac{1}{2}\times 50\left(x+0.2\right)^{2}$$

Multiply \frac{1}{2} and 50 to get 25.

$$5=125x^{2}+25\left(x+0.2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+0.2\right)^{2}.

$$5=125x^{2}+25\left(x^{2}+0.4x+0.04\right)$$

Use the distributive property to multiply 25 by x^{2}+0.4x+0.04.

$$5=125x^{2}+25x^{2}+10x+1$$

Combine 125x^{2} and 25x^{2} to get 150x^{2}.

$$5=150x^{2}+10x+1$$

Swap sides so that all variable terms are on the left hand side.

$$150x^{2}+10x+1=5$$

Subtract 5 from both sides.

$$150x^{2}+10x+1-5=0$$

Subtract 5 from 1 to get -4.

$$150x^{2}+10x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 150 for a, 10 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 150\left(-4\right)}}{2\times 150}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 150\left(-4\right)}}{2\times 150}$$

Multiply -4 times 150.

$$x=\frac{-10±\sqrt{100-600\left(-4\right)}}{2\times 150}$$

Multiply -600 times -4.

$$x=\frac{-10±\sqrt{100+2400}}{2\times 150}$$

Add 100 to 2400.

$$x=\frac{-10±\sqrt{2500}}{2\times 150}$$

Take the square root of 2500.

$$x=\frac{-10±50}{2\times 150}$$

Multiply 2 times 150.

$$x=\frac{-10±50}{300}$$

Now solve the equation x=\frac{-10±50}{300} when ± is plus. Add -10 to 50.

$$x=\frac{40}{300}$$

Reduce the fraction \frac{40}{300} to lowest terms by extracting and canceling out 20.

$$x=\frac{2}{15}$$

Now solve the equation x=\frac{-10±50}{300} when ± is minus. Subtract 50 from -10.

$$x=-\frac{60}{300}$$

Reduce the fraction \frac{-60}{300} to lowest terms by extracting and canceling out 60.

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

The equation is now solved.

$$x=\frac{2}{15} x=-\frac{1}{5}$$

Steps for Completing the Square

Multiply \frac{1}{2} and 250 to get 125.

$$5=125x^{2}+\frac{1}{2}\times 50\left(x+0.2\right)^{2}$$

Multiply \frac{1}{2} and 50 to get 25.

$$5=125x^{2}+25\left(x+0.2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+0.2\right)^{2}.

$$5=125x^{2}+25\left(x^{2}+0.4x+0.04\right)$$

Use the distributive property to multiply 25 by x^{2}+0.4x+0.04.

$$5=125x^{2}+25x^{2}+10x+1$$

Combine 125x^{2} and 25x^{2} to get 150x^{2}.

$$5=150x^{2}+10x+1$$

Swap sides so that all variable terms are on the left hand side.

$$150x^{2}+10x+1=5$$

Subtract 1 from both sides.

$$150x^{2}+10x=5-1$$

Subtract 1 from 5 to get 4.

$$150x^{2}+10x=4$$

Divide both sides by 150.

$$\frac{150x^{2}+10x}{150}=\frac{4}{150}$$

Dividing by 150 undoes the multiplication by 150.

$$x^{2}+\frac{10}{150}x=\frac{4}{150}$$

Reduce the fraction \frac{10}{150} to lowest terms by extracting and canceling out 10.

$$x^{2}+\frac{1}{15}x=\frac{4}{150}$$

Reduce the fraction \frac{4}{150} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{15}x=\frac{2}{75}$$

Divide \frac{1}{15}, the coefficient of the x term, by 2 to get \frac{1}{30}. Then add the square of \frac{1}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{15}x+\left(\frac{1}{30}\right)^{2}=\frac{2}{75}+\left(\frac{1}{30}\right)^{2}$$

Square \frac{1}{30} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{15}x+\frac{1}{900}=\frac{2}{75}+\frac{1}{900}$$

Add \frac{2}{75} to \frac{1}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{15}x+\frac{1}{900}=\frac{1}{36}$$

Factor x^{2}+\frac{1}{15}x+\frac{1}{900}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{30}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{30}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{1}{30}=\frac{1}{6} x+\frac{1}{30}=-\frac{1}{6}$$

Subtract \frac{1}{30} from both sides of the equation.

$$x=\frac{2}{15} x=-\frac{1}{5}$$