Solve 5=1/2*20*x^2+1/2*50(x+0.2)^2

$$5 = \frac { 1 } { 2 } \cdot 20 \cdot x ^ { 2 } + \frac { 1 } { 2 } \cdot 50 ( x + 0.2 ) ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{165}}{35}-\frac{1}{7}\approx 0.224149502$$

$$x=-\frac{\sqrt{165}}{35}-\frac{1}{7}\approx -0.509863788$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 20 to get 10.

$$5=10x^{2}+\frac{1}{2}\times 50\left(x+0.2\right)^{2}$$

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

$$5=10x^{2}+25\left(x+0.2\right)^{2}$$

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

$$5=10x^{2}+25\left(x^{2}+0.4x+0.04\right)$$

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

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

Combine 10x^{2} and 25x^{2} to get 35x^{2}.

$$5=35x^{2}+10x+1$$

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

$$35x^{2}+10x+1=5$$

Subtract 5 from both sides.

$$35x^{2}+10x+1-5=0$$

Subtract 5 from 1 to get -4.

$$35x^{2}+10x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 35 for a, 10 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 35\left(-4\right)}}{2\times 35}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 35\left(-4\right)}}{2\times 35}$$

Multiply -4 times 35.

$$x=\frac{-10±\sqrt{100-140\left(-4\right)}}{2\times 35}$$

Multiply -140 times -4.

$$x=\frac{-10±\sqrt{100+560}}{2\times 35}$$

Add 100 to 560.

$$x=\frac{-10±\sqrt{660}}{2\times 35}$$

Take the square root of 660.

$$x=\frac{-10±2\sqrt{165}}{2\times 35}$$

Multiply 2 times 35.

$$x=\frac{-10±2\sqrt{165}}{70}$$

Now solve the equation x=\frac{-10±2\sqrt{165}}{70} when ± is plus. Add -10 to 2\sqrt{165}.

$$x=\frac{2\sqrt{165}-10}{70}$$

Divide -10+2\sqrt{165} by 70.

$$x=\frac{\sqrt{165}}{35}-\frac{1}{7}$$

Now solve the equation x=\frac{-10±2\sqrt{165}}{70} when ± is minus. Subtract 2\sqrt{165} from -10.

$$x=\frac{-2\sqrt{165}-10}{70}$$

Divide -10-2\sqrt{165} by 70.

$$x=-\frac{\sqrt{165}}{35}-\frac{1}{7}$$

The equation is now solved.

$$x=\frac{\sqrt{165}}{35}-\frac{1}{7} x=-\frac{\sqrt{165}}{35}-\frac{1}{7}$$

Steps for Completing the Square

Multiply \frac{1}{2} and 20 to get 10.

$$5=10x^{2}+\frac{1}{2}\times 50\left(x+0.2\right)^{2}$$

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

$$5=10x^{2}+25\left(x+0.2\right)^{2}$$

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

$$5=10x^{2}+25\left(x^{2}+0.4x+0.04\right)$$

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

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

Combine 10x^{2} and 25x^{2} to get 35x^{2}.

$$5=35x^{2}+10x+1$$

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

$$35x^{2}+10x+1=5$$

Subtract 1 from both sides.

$$35x^{2}+10x=5-1$$

Subtract 1 from 5 to get 4.

$$35x^{2}+10x=4$$

Divide both sides by 35.

$$\frac{35x^{2}+10x}{35}=\frac{4}{35}$$

Dividing by 35 undoes the multiplication by 35.

$$x^{2}+\frac{10}{35}x=\frac{4}{35}$$

Reduce the fraction \frac{10}{35} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{2}{7}x=\frac{4}{35}$$

Divide \frac{2}{7}, the coefficient of the x term, by 2 to get \frac{1}{7}. Then add the square of \frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{7}x+\left(\frac{1}{7}\right)^{2}=\frac{4}{35}+\left(\frac{1}{7}\right)^{2}$$

Square \frac{1}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{7}x+\frac{1}{49}=\frac{4}{35}+\frac{1}{49}$$

Add \frac{4}{35} to \frac{1}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{7}x+\frac{1}{49}=\frac{33}{245}$$

Factor x^{2}+\frac{2}{7}x+\frac{1}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{7}\right)^{2}=\frac{33}{245}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{7}\right)^{2}}=\sqrt{\frac{33}{245}}$$

Simplify.

$$x+\frac{1}{7}=\frac{\sqrt{165}}{35} x+\frac{1}{7}=-\frac{\sqrt{165}}{35}$$

Subtract \frac{1}{7} from both sides of the equation.

$$x=\frac{\sqrt{165}}{35}-\frac{1}{7} x=-\frac{\sqrt{165}}{35}-\frac{1}{7}$$