Solve left(2x-5right)^2+x^2+6(2x-5)-12x+20=0

$${ \left(2x-5 \right) }^{ 2 } + { x }^{ 2 } +6(2x-5)-12x+20=0$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$4x^{2}-20x+25+x^{2}+6\left(2x-5\right)-12x+20=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-20x+25+6\left(2x-5\right)-12x+20=0$$

Use the distributive property to multiply 6 by 2x-5.

$$5x^{2}-20x+25+12x-30-12x+20=0$$

Combine -20x and 12x to get -8x.

$$5x^{2}-8x+25-30-12x+20=0$$

Subtract 30 from 25 to get -5.

$$5x^{2}-8x-5-12x+20=0$$

Combine -8x and -12x to get -20x.

$$5x^{2}-20x-5+20=0$$

Add -5 and 20 to get 15.

$$5x^{2}-20x+15=0$$

Divide both sides by 5.

$$x^{2}-4x+3=0$$

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

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite x^{2}-4x+3 as \left(x^{2}-3x\right)+\left(-x+3\right).

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

Factor out x in the first and -1 in the second group.

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

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-1\right)$$

To find equation solutions, solve x-3=0 and x-1=0.

$$x=3 x=1$$

Steps Using the Quadratic Formula

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

$$4x^{2}-20x+25+x^{2}+6\left(2x-5\right)-12x+20=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-20x+25+6\left(2x-5\right)-12x+20=0$$

Use the distributive property to multiply 6 by 2x-5.

$$5x^{2}-20x+25+12x-30-12x+20=0$$

Combine -20x and 12x to get -8x.

$$5x^{2}-8x+25-30-12x+20=0$$

Subtract 30 from 25 to get -5.

$$5x^{2}-8x-5-12x+20=0$$

Combine -8x and -12x to get -20x.

$$5x^{2}-20x-5+20=0$$

Add -5 and 20 to get 15.

$$5x^{2}-20x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -20 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 5\times 15}}{2\times 5}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 5\times 15}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-20\right)±\sqrt{400-20\times 15}}{2\times 5}$$

Multiply -20 times 15.

$$x=\frac{-\left(-20\right)±\sqrt{400-300}}{2\times 5}$$

Add 400 to -300.

$$x=\frac{-\left(-20\right)±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$x=\frac{-\left(-20\right)±10}{2\times 5}$$

The opposite of -20 is 20.

$$x=\frac{20±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{20±10}{10}$$

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

$$x=\frac{30}{10}$$

Divide 30 by 10.

$$x=3$$

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

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

Divide 10 by 10.

$$x=1$$

The equation is now solved.

$$x=3 x=1$$

Steps for Completing the Square

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

$$4x^{2}-20x+25+x^{2}+6\left(2x-5\right)-12x+20=0$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$5x^{2}-20x+25+6\left(2x-5\right)-12x+20=0$$

Use the distributive property to multiply 6 by 2x-5.

$$5x^{2}-20x+25+12x-30-12x+20=0$$

Combine -20x and 12x to get -8x.

$$5x^{2}-8x+25-30-12x+20=0$$

Subtract 30 from 25 to get -5.

$$5x^{2}-8x-5-12x+20=0$$

Combine -8x and -12x to get -20x.

$$5x^{2}-20x-5+20=0$$

Add -5 and 20 to get 15.

$$5x^{2}-20x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-20x=-15$$

Divide both sides by 5.

$$\frac{5x^{2}-20x}{5}=-\frac{15}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{20}{5}\right)x=-\frac{15}{5}$$

Divide -20 by 5.

$$x^{2}-4x=-\frac{15}{5}$$

Divide -15 by 5.

$$x^{2}-4x=-3$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-3+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

$$x^{2}-4x+4=1$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$