Solve 2^{2}+4^{2}+6^2+8^2+x^2=5*(4+x/5)^2+5*4

$$2 ^ { 2 } + 4 ^ { 2 } + 6 ^ { 2 } + 8 ^ { 2 } + x ^ { 2 } = 5 \times ( 4 + \frac { x } { 5 } ) ^ { 2 } + 5 \times 4$$

Solve for x

$$x=5$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 2 and get 4.

$$4+4^{2}+6^{2}+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Calculate 4 to the power of 2 and get 16.

$$4+16+6^{2}+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Add 4 and 16 to get 20.

$$20+6^{2}+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Calculate 6 to the power of 2 and get 36.

$$20+36+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Add 20 and 36 to get 56.

$$56+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Calculate 8 to the power of 2 and get 64.

$$56+64+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Add 56 and 64 to get 120.

$$120+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+\frac{x}{5}\right)^{2}.

$$120+x^{2}=5\left(16+8\times \frac{x}{5}+\left(\frac{x}{5}\right)^{2}\right)+5\times 4$$

Express 8\times \frac{x}{5} as a single fraction.

$$120+x^{2}=5\left(16+\frac{8x}{5}+\left(\frac{x}{5}\right)^{2}\right)+5\times 4$$

To raise \frac{x}{5} to a power, raise both numerator and denominator to the power and then divide.

$$120+x^{2}=5\left(16+\frac{8x}{5}+\frac{x^{2}}{5^{2}}\right)+5\times 4$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{5^{2}}{5^{2}}.

$$120+x^{2}=5\left(\frac{16\times 5^{2}}{5^{2}}+\frac{8x}{5}+\frac{x^{2}}{5^{2}}\right)+5\times 4$$

Since \frac{16\times 5^{2}}{5^{2}} and \frac{x^{2}}{5^{2}} have the same denominator, add them by adding their numerators.

$$120+x^{2}=5\left(\frac{16\times 5^{2}+x^{2}}{5^{2}}+\frac{8x}{5}\right)+5\times 4$$

Do the multiplications in 16\times 5^{2}+x^{2}.

$$120+x^{2}=5\left(\frac{400+x^{2}}{5^{2}}+\frac{8x}{5}\right)+5\times 4$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5^{2} and 5 is 25. Multiply \frac{8x}{5} times \frac{5}{5}.

$$120+x^{2}=5\left(\frac{400+x^{2}}{25}+\frac{5\times 8x}{25}\right)+5\times 4$$

Since \frac{400+x^{2}}{25} and \frac{5\times 8x}{25} have the same denominator, add them by adding their numerators.

$$120+x^{2}=5\times \frac{400+x^{2}+5\times 8x}{25}+5\times 4$$

Do the multiplications in 400+x^{2}+5\times 8x.

$$120+x^{2}=5\times \frac{400+x^{2}+40x}{25}+5\times 4$$

Cancel out 25, the greatest common factor in 5 and 25.

$$120+x^{2}=\frac{400+x^{2}+40x}{5}+5\times 4$$

Multiply 5 and 4 to get 20.

$$120+x^{2}=\frac{400+x^{2}+40x}{5}+20$$

Divide each term of 400+x^{2}+40x by 5 to get 80+\frac{1}{5}x^{2}+8x.

$$120+x^{2}=80+\frac{1}{5}x^{2}+8x+20$$

Add 80 and 20 to get 100.

$$120+x^{2}=100+\frac{1}{5}x^{2}+8x$$

Subtract 100 from both sides.

$$120+x^{2}-100=\frac{1}{5}x^{2}+8x$$

Subtract 100 from 120 to get 20.

$$20+x^{2}=\frac{1}{5}x^{2}+8x$$

Subtract \frac{1}{5}x^{2} from both sides.

$$20+x^{2}-\frac{1}{5}x^{2}=8x$$

Combine x^{2} and -\frac{1}{5}x^{2} to get \frac{4}{5}x^{2}.

$$20+\frac{4}{5}x^{2}=8x$$

Subtract 8x from both sides.

$$20+\frac{4}{5}x^{2}-8x=0$$

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.

$$\frac{4}{5}x^{2}-8x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{4}{5} for a, -8 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times \frac{4}{5}\times 20}}{2\times \frac{4}{5}}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times \frac{4}{5}\times 20}}{2\times \frac{4}{5}}$$

Multiply -4 times \frac{4}{5}.

$$x=\frac{-\left(-8\right)±\sqrt{64-\frac{16}{5}\times 20}}{2\times \frac{4}{5}}$$

Multiply -\frac{16}{5} times 20.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2\times \frac{4}{5}}$$

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2\times \frac{4}{5}}$$

Take the square root of 0.

$$x=-\frac{-8}{2\times \frac{4}{5}}$$

The opposite of -8 is 8.

$$x=\frac{8}{2\times \frac{4}{5}}$$

Multiply 2 times \frac{4}{5}.

$$x=\frac{8}{\frac{8}{5}}$$

Divide 8 by \frac{8}{5} by multiplying 8 by the reciprocal of \frac{8}{5}.

$$x=5$$

Steps for Completing the Square

Calculate 2 to the power of 2 and get 4.

$$4+4^{2}+6^{2}+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Calculate 4 to the power of 2 and get 16.

$$4+16+6^{2}+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Add 4 and 16 to get 20.

$$20+6^{2}+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Calculate 6 to the power of 2 and get 36.

$$20+36+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Add 20 and 36 to get 56.

$$56+8^{2}+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Calculate 8 to the power of 2 and get 64.

$$56+64+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Add 56 and 64 to get 120.

$$120+x^{2}=5\left(4+\frac{x}{5}\right)^{2}+5\times 4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+\frac{x}{5}\right)^{2}.

$$120+x^{2}=5\left(16+8\times \frac{x}{5}+\left(\frac{x}{5}\right)^{2}\right)+5\times 4$$

Express 8\times \frac{x}{5} as a single fraction.

$$120+x^{2}=5\left(16+\frac{8x}{5}+\left(\frac{x}{5}\right)^{2}\right)+5\times 4$$

To raise \frac{x}{5} to a power, raise both numerator and denominator to the power and then divide.

$$120+x^{2}=5\left(16+\frac{8x}{5}+\frac{x^{2}}{5^{2}}\right)+5\times 4$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 16 times \frac{5^{2}}{5^{2}}.

$$120+x^{2}=5\left(\frac{16\times 5^{2}}{5^{2}}+\frac{8x}{5}+\frac{x^{2}}{5^{2}}\right)+5\times 4$$

Since \frac{16\times 5^{2}}{5^{2}} and \frac{x^{2}}{5^{2}} have the same denominator, add them by adding their numerators.

$$120+x^{2}=5\left(\frac{16\times 5^{2}+x^{2}}{5^{2}}+\frac{8x}{5}\right)+5\times 4$$

Do the multiplications in 16\times 5^{2}+x^{2}.

$$120+x^{2}=5\left(\frac{400+x^{2}}{5^{2}}+\frac{8x}{5}\right)+5\times 4$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5^{2} and 5 is 25. Multiply \frac{8x}{5} times \frac{5}{5}.

$$120+x^{2}=5\left(\frac{400+x^{2}}{25}+\frac{5\times 8x}{25}\right)+5\times 4$$

Since \frac{400+x^{2}}{25} and \frac{5\times 8x}{25} have the same denominator, add them by adding their numerators.

$$120+x^{2}=5\times \frac{400+x^{2}+5\times 8x}{25}+5\times 4$$

Do the multiplications in 400+x^{2}+5\times 8x.

$$120+x^{2}=5\times \frac{400+x^{2}+40x}{25}+5\times 4$$

Cancel out 25, the greatest common factor in 5 and 25.

$$120+x^{2}=\frac{400+x^{2}+40x}{5}+5\times 4$$

Multiply 5 and 4 to get 20.

$$120+x^{2}=\frac{400+x^{2}+40x}{5}+20$$

Divide each term of 400+x^{2}+40x by 5 to get 80+\frac{1}{5}x^{2}+8x.

$$120+x^{2}=80+\frac{1}{5}x^{2}+8x+20$$

Add 80 and 20 to get 100.

$$120+x^{2}=100+\frac{1}{5}x^{2}+8x$$

Subtract \frac{1}{5}x^{2} from both sides.

$$120+x^{2}-\frac{1}{5}x^{2}=100+8x$$

Combine x^{2} and -\frac{1}{5}x^{2} to get \frac{4}{5}x^{2}.

$$120+\frac{4}{5}x^{2}=100+8x$$

Subtract 8x from both sides.

$$120+\frac{4}{5}x^{2}-8x=100$$

Subtract 120 from both sides.

$$\frac{4}{5}x^{2}-8x=100-120$$

Subtract 120 from 100 to get -20.

$$\frac{4}{5}x^{2}-8x=-20$$

Divide both sides of the equation by \frac{4}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{4}{5}x^{2}-8x}{\frac{4}{5}}=-\frac{20}{\frac{4}{5}}$$

Dividing by \frac{4}{5} undoes the multiplication by \frac{4}{5}.

$$x^{2}+\left(-\frac{8}{\frac{4}{5}}\right)x=-\frac{20}{\frac{4}{5}}$$

Divide -8 by \frac{4}{5} by multiplying -8 by the reciprocal of \frac{4}{5}.

$$x^{2}-10x=-\frac{20}{\frac{4}{5}}$$

Divide -20 by \frac{4}{5} by multiplying -20 by the reciprocal of \frac{4}{5}.

$$x^{2}-10x=-25$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -5.

$$x^{2}-10x+25=-25+25$$

Add -25 to 25.

$$x^{2}-10x+25=0$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$