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

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

Solve for x

$$x=2$$

$$x = \frac{14}{3} = 4\frac{2}{3} \approx 4.666666667$$

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}.

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

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

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

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

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

Subtract 3 from both sides.

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

Subtract 3 from -25 to get -28.

$$-3x^{2}-28+20x=0$$

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

$$-3x^{2}+20x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=20 ab=-3\left(-28\right)=84$$

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 84.

$$1,84 2,42 3,28 4,21 6,14 7,12$$

Calculate the sum for each pair.

$$1+84=85 2+42=44 3+28=31 4+21=25 6+14=20 7+12=19$$

The solution is the pair that gives sum 20.

$$a=14 b=6$$

Rewrite -3x^{2}+20x-28 as \left(-3x^{2}+14x\right)+\left(6x-28\right).

$$\left(-3x^{2}+14x\right)+\left(6x-28\right)$$

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

$$-x\left(3x-14\right)+2\left(3x-14\right)$$

Factor out common term 3x-14 by using distributive property.

$$\left(3x-14\right)\left(-x+2\right)$$

To find equation solutions, solve 3x-14=0 and -x+2=0.

$$x=\frac{14}{3} x=2$$

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}.

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

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

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

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

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

Subtract 3 from both sides.

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

Subtract 3 from -25 to get -28.

$$-3x^{2}-28+20x=0$$

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.

$$-3x^{2}+20x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 20 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-3\right)\left(-28\right)}}{2\left(-3\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-3\right)\left(-28\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-20±\sqrt{400+12\left(-28\right)}}{2\left(-3\right)}$$

Multiply 12 times -28.

$$x=\frac{-20±\sqrt{400-336}}{2\left(-3\right)}$$

Add 400 to -336.

$$x=\frac{-20±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-20±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-20±8}{-6}$$

Now solve the equation x=\frac{-20±8}{-6} when ± is plus. Add -20 to 8.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

Now solve the equation x=\frac{-20±8}{-6} when ± is minus. Subtract 8 from -20.

$$x=-\frac{28}{-6}$$

Reduce the fraction \frac{-28}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{14}{3}$$

The equation is now solved.

$$x=2 x=\frac{14}{3}$$

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}.

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

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

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

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

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

Add 25 to both sides.

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

Add 3 and 25 to get 28.

$$-3x^{2}+20x=28$$

Divide both sides by -3.

$$\frac{-3x^{2}+20x}{-3}=\frac{28}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{20}{-3}x=\frac{28}{-3}$$

Divide 20 by -3.

$$x^{2}-\frac{20}{3}x=\frac{28}{-3}$$

Divide 28 by -3.

$$x^{2}-\frac{20}{3}x=-\frac{28}{3}$$

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.

$$x^{2}-\frac{20}{3}x+\left(-\frac{10}{3}\right)^{2}=-\frac{28}{3}+\left(-\frac{10}{3}\right)^{2}$$

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

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=-\frac{28}{3}+\frac{100}{9}$$

Add -\frac{28}{3} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{3}x+\frac{100}{9}=\frac{16}{9}$$

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}.

$$\left(x-\frac{10}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{10}{3}=\frac{4}{3} x-\frac{10}{3}=-\frac{4}{3}$$

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

$$x=\frac{14}{3} x=2$$