Solve 3x^2-left(5-2xright)^2=11

$$3 { x }^{ 2 } - { \left(5-2x \right) }^{ 2 } =11$$

Solve for x

$$x=2$$

$$x=18$$

Steps Using Factoring By Grouping

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

$$3x^{2}-\left(25-20x+4x^{2}\right)=11$$

To find the opposite of 25-20x+4x^{2}, find the opposite of each term.

$$3x^{2}-25+20x-4x^{2}=11$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-25+20x=11$$

Subtract 11 from both sides.

$$-x^{2}-25+20x-11=0$$

Subtract 11 from -25 to get -36.

$$-x^{2}-36+20x=0$$

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

$$-x^{2}+20x-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-36. To find a and b, set up a system to be solved.

$$a+b=20 ab=-\left(-36\right)=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 20.

$$a=18 b=2$$

Rewrite -x^{2}+20x-36 as \left(-x^{2}+18x\right)+\left(2x-36\right).

$$\left(-x^{2}+18x\right)+\left(2x-36\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-18\right)+2\left(x-18\right)$$

Factor out common term x-18 by using distributive property.

$$\left(x-18\right)\left(-x+2\right)$$

To find equation solutions, solve x-18=0 and -x+2=0.

$$x=18 x=2$$

Steps Using the Quadratic Formula

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

$$3x^{2}-\left(25-20x+4x^{2}\right)=11$$

To find the opposite of 25-20x+4x^{2}, find the opposite of each term.

$$3x^{2}-25+20x-4x^{2}=11$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-25+20x=11$$

Subtract 11 from both sides.

$$-x^{2}-25+20x-11=0$$

Subtract 11 from -25 to get -36.

$$-x^{2}-36+20x=0$$

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.

$$-x^{2}+20x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 20 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-1\right)\left(-36\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-20±\sqrt{400+4\left(-36\right)}}{2\left(-1\right)}$$

Multiply 4 times -36.

$$x=\frac{-20±\sqrt{400-144}}{2\left(-1\right)}$$

Add 400 to -144.

$$x=\frac{-20±\sqrt{256}}{2\left(-1\right)}$$

Take the square root of 256.

$$x=\frac{-20±16}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-20±16}{-2}$$

Now solve the equation x=\frac{-20±16}{-2} when ± is plus. Add -20 to 16.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-20±16}{-2} when ± is minus. Subtract 16 from -20.

$$x=-\frac{36}{-2}$$

Divide -36 by -2.

$$x=18$$

The equation is now solved.

$$x=2 x=18$$

Steps for Completing the Square

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

$$3x^{2}-\left(25-20x+4x^{2}\right)=11$$

To find the opposite of 25-20x+4x^{2}, find the opposite of each term.

$$3x^{2}-25+20x-4x^{2}=11$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-25+20x=11$$

Add 25 to both sides.

$$-x^{2}+20x=11+25$$

Add 11 and 25 to get 36.

$$-x^{2}+20x=36$$

Divide both sides by -1.

$$\frac{-x^{2}+20x}{-1}=\frac{36}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{20}{-1}x=\frac{36}{-1}$$

Divide 20 by -1.

$$x^{2}-20x=\frac{36}{-1}$$

Divide 36 by -1.

$$x^{2}-20x=-36$$

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

$$x^{2}-20x+\left(-10\right)^{2}=-36+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-36+100$$

Add -36 to 100.

$$x^{2}-20x+100=64$$

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

$$\left(x-10\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{64}$$

Simplify.

$$x-10=8 x-10=-8$$

Add 10 to both sides of the equation.

$$x=18 x=2$$