Solve x^2+left(-3x+10right)^2=20

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

Solve for x

$$x=2$$

$$x=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-3x+10\right)^{2}.

$$x^{2}+9x^{2}-60x+100=20$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}-60x+100=20$$

Subtract 20 from both sides.

$$10x^{2}-60x+100-20=0$$

Subtract 20 from 100 to get 80.

$$10x^{2}-60x+80=0$$

Divide both sides by 10.

$$x^{2}-6x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite x^{2}-6x+8 as \left(x^{2}-4x\right)+\left(-2x+8\right).

$$\left(x^{2}-4x\right)+\left(-2x+8\right)$$

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

$$x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-2\right)$$

To find equation solutions, solve x-4=0 and x-2=0.

$$x=4 x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-3x+10\right)^{2}.

$$x^{2}+9x^{2}-60x+100=20$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}-60x+100=20$$

Subtract 20 from both sides.

$$10x^{2}-60x+100-20=0$$

Subtract 20 from 100 to get 80.

$$10x^{2}-60x+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -60 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 10\times 80}}{2\times 10}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 10\times 80}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-60\right)±\sqrt{3600-40\times 80}}{2\times 10}$$

Multiply -40 times 80.

$$x=\frac{-\left(-60\right)±\sqrt{3600-3200}}{2\times 10}$$

Add 3600 to -3200.

$$x=\frac{-\left(-60\right)±\sqrt{400}}{2\times 10}$$

Take the square root of 400.

$$x=\frac{-\left(-60\right)±20}{2\times 10}$$

The opposite of -60 is 60.

$$x=\frac{60±20}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{60±20}{20}$$

Now solve the equation x=\frac{60±20}{20} when ± is plus. Add 60 to 20.

$$x=\frac{80}{20}$$

Divide 80 by 20.

$$x=4$$

Now solve the equation x=\frac{60±20}{20} when ± is minus. Subtract 20 from 60.

$$x=\frac{40}{20}$$

Divide 40 by 20.

$$x=2$$

The equation is now solved.

$$x=4 x=2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-3x+10\right)^{2}.

$$x^{2}+9x^{2}-60x+100=20$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}-60x+100=20$$

Subtract 100 from both sides.

$$10x^{2}-60x=20-100$$

Subtract 100 from 20 to get -80.

$$10x^{2}-60x=-80$$

Divide both sides by 10.

$$\frac{10x^{2}-60x}{10}=-\frac{80}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{60}{10}\right)x=-\frac{80}{10}$$

Divide -60 by 10.

$$x^{2}-6x=-\frac{80}{10}$$

Divide -80 by 10.

$$x^{2}-6x=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-8+9$$

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-3=1 x-3=-1$$

Add 3 to both sides of the equation.

$$x=4 x=2$$