Solve (x-5)^2=(x+5)^2-3x^2-25

$$( x - 5 ) ^ { 2 } = ( x + 5 ) ^ { 2 } - 3 x ^ { 2 } - 25$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=5$$

Steps Using Factoring By Grouping

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

$$x^{2}-10x+25=\left(x+5\right)^{2}-3x^{2}-25$$

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

$$x^{2}-10x+25=x^{2}+10x+25-3x^{2}-25$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

Subtract 25 from 25 to get 0.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-10x+25=10x$$

Subtract 10x from both sides.

$$3x^{2}-10x+25-10x=0$$

Combine -10x and -10x to get -20x.

$$3x^{2}-20x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\times 25=75$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 75.

$$-1,-75 -3,-25 -5,-15$$

Calculate the sum for each pair.

$$-1-75=-76 -3-25=-28 -5-15=-20$$

The solution is the pair that gives sum -20.

$$a=-15 b=-5$$

Rewrite 3x^{2}-20x+25 as \left(3x^{2}-15x\right)+\left(-5x+25\right).

$$\left(3x^{2}-15x\right)+\left(-5x+25\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(x-5\right)-5\left(x-5\right)$$

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

$$\left(x-5\right)\left(3x-5\right)$$

To find equation solutions, solve x-5=0 and 3x-5=0.

$$x=5 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

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

$$x^{2}-10x+25=\left(x+5\right)^{2}-3x^{2}-25$$

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

$$x^{2}-10x+25=x^{2}+10x+25-3x^{2}-25$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

Subtract 25 from 25 to get 0.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-10x+25=10x$$

Subtract 10x from both sides.

$$3x^{2}-10x+25-10x=0$$

Combine -10x and -10x to get -20x.

$$3x^{2}-20x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -20 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\times 25}}{2\times 3}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\times 25}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\times 25}}{2\times 3}$$

Multiply -12 times 25.

$$x=\frac{-\left(-20\right)±\sqrt{400-300}}{2\times 3}$$

Add 400 to -300.

$$x=\frac{-\left(-20\right)±\sqrt{100}}{2\times 3}$$

Take the square root of 100.

$$x=\frac{-\left(-20\right)±10}{2\times 3}$$

The opposite of -20 is 20.

$$x=\frac{20±10}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{20±10}{6}$$

Now solve the equation x=\frac{20±10}{6} when ± is plus. Add 20 to 10.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{20±10}{6} when ± is minus. Subtract 10 from 20.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

The equation is now solved.

$$x=5 x=\frac{5}{3}$$

Steps for Completing the Square

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

$$x^{2}-10x+25=\left(x+5\right)^{2}-3x^{2}-25$$

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

$$x^{2}-10x+25=x^{2}+10x+25-3x^{2}-25$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

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

Subtract 25 from 25 to get 0.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-10x+25=10x$$

Subtract 10x from both sides.

$$3x^{2}-10x+25-10x=0$$

Combine -10x and -10x to get -20x.

$$3x^{2}-20x+25=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-20x=-25$$

Divide both sides by 3.

$$\frac{3x^{2}-20x}{3}=-\frac{25}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{20}{3}x=-\frac{25}{3}$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=-\frac{25}{3}+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=-\frac{25}{3}+\frac{100}{9}$$

Add -\frac{25}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{25}{9}$$

Factor x^{2}-\frac{20}{3}x+\frac{100}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{5}{3} x-\frac{10}{3}=-\frac{5}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$x=5 x=\frac{5}{3}$$