Solve 14^2=10^2+x^2-2*10*x*1/2

$${ 14 }^{ 2 } = { 10 }^{ 2 } + { x }^{ 2 } -2 \times 10 \times x \times \frac{ 1 }{ 2 }$$

Solve for x

$$x=-6$$

$$x=16$$

Steps Using Factoring

Calculate 14 to the power of 2 and get 196.

$$196=10^{2}+x^{2}-2\times 10x\times \frac{1}{2}$$

Calculate 10 to the power of 2 and get 100.

$$196=100+x^{2}-2\times 10x\times \frac{1}{2}$$

Multiply 2 and 10 to get 20.

$$196=100+x^{2}-20x\times \frac{1}{2}$$

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

$$196=100+x^{2}-10x$$

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

$$100+x^{2}-10x=196$$

Subtract 196 from both sides.

$$100+x^{2}-10x-196=0$$

Subtract 196 from 100 to get -96.

$$-96+x^{2}-10x=0$$

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

$$x^{2}-10x-96=0$$

To solve the equation, factor x^{2}-10x-96 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=-96$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -10.

$$a=-16 b=6$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-16\right)\left(x+6\right)$$

To find equation solutions, solve x-16=0 and x+6=0.

$$x=16 x=-6$$

Steps Using Factoring By Grouping

Calculate 14 to the power of 2 and get 196.

$$196=10^{2}+x^{2}-2\times 10x\times \frac{1}{2}$$

Calculate 10 to the power of 2 and get 100.

$$196=100+x^{2}-2\times 10x\times \frac{1}{2}$$

Multiply 2 and 10 to get 20.

$$196=100+x^{2}-20x\times \frac{1}{2}$$

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

$$196=100+x^{2}-10x$$

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

$$100+x^{2}-10x=196$$

Subtract 196 from both sides.

$$100+x^{2}-10x-196=0$$

Subtract 196 from 100 to get -96.

$$-96+x^{2}-10x=0$$

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

$$x^{2}-10x-96=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-96. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-96\right)=-96$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -10.

$$a=-16 b=6$$

Rewrite x^{2}-10x-96 as \left(x^{2}-16x\right)+\left(6x-96\right).

$$\left(x^{2}-16x\right)+\left(6x-96\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-16\right)+6\left(x-16\right)$$

Factor out common term x-16 by using distributive property.

$$\left(x-16\right)\left(x+6\right)$$

To find equation solutions, solve x-16=0 and x+6=0.

$$x=16 x=-6$$

Steps Using the Quadratic Formula

Calculate 14 to the power of 2 and get 196.

$$196=10^{2}+x^{2}-2\times 10x\times \frac{1}{2}$$

Calculate 10 to the power of 2 and get 100.

$$196=100+x^{2}-2\times 10x\times \frac{1}{2}$$

Multiply 2 and 10 to get 20.

$$196=100+x^{2}-20x\times \frac{1}{2}$$

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

$$196=100+x^{2}-10x$$

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

$$100+x^{2}-10x=196$$

Subtract 196 from both sides.

$$100+x^{2}-10x-196=0$$

Subtract 196 from 100 to get -96.

$$-96+x^{2}-10x=0$$

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.

$$x^{2}-10x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-96\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-96\right)}}{2}$$

Multiply -4 times -96.

$$x=\frac{-\left(-10\right)±\sqrt{100+384}}{2}$$

Add 100 to 384.

$$x=\frac{-\left(-10\right)±\sqrt{484}}{2}$$

Take the square root of 484.

$$x=\frac{-\left(-10\right)±22}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±22}{2}$$

Now solve the equation x=\frac{10±22}{2} when ± is plus. Add 10 to 22.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Now solve the equation x=\frac{10±22}{2} when ± is minus. Subtract 22 from 10.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=16 x=-6$$

Steps for Completing the Square

Calculate 14 to the power of 2 and get 196.

$$196=10^{2}+x^{2}-2\times 10x\times \frac{1}{2}$$

Calculate 10 to the power of 2 and get 100.

$$196=100+x^{2}-2\times 10x\times \frac{1}{2}$$

Multiply 2 and 10 to get 20.

$$196=100+x^{2}-20x\times \frac{1}{2}$$

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

$$196=100+x^{2}-10x$$

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

$$100+x^{2}-10x=196$$

Subtract 100 from both sides.

$$x^{2}-10x=196-100$$

Subtract 100 from 196 to get 96.

$$x^{2}-10x=96$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=96+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=96+25$$

Add 96 to 25.

$$x^{2}-10x+25=121$$

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}.

$$\left(x-5\right)^{2}=121$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{121}$$

Simplify.

$$x-5=11 x-5=-11$$

Add 5 to both sides of the equation.

$$x=16 x=-6$$