Solve left(x-10right)^2=10*(70-x)

$${ \left(x-10 \right) }^{ 2 } =10 \times (70-x)$$

Solve for x

$$x=-20$$

$$x=30$$

Steps Using Factoring

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

$$x^{2}-20x+100=10\left(70-x\right)$$

Use the distributive property to multiply 10 by 70-x.

$$x^{2}-20x+100=700-10x$$

Subtract 700 from both sides.

$$x^{2}-20x+100-700=-10x$$

Subtract 700 from 100 to get -600.

$$x^{2}-20x-600=-10x$$

Add 10x to both sides.

$$x^{2}-20x-600+10x=0$$

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

$$x^{2}-10x-600=0$$

To solve the equation, factor x^{2}-10x-600 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=-600$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -600.

$$1,-600 2,-300 3,-200 4,-150 5,-120 6,-100 8,-75 10,-60 12,-50 15,-40 20,-30 24,-25$$

Calculate the sum for each pair.

$$1-600=-599 2-300=-298 3-200=-197 4-150=-146 5-120=-115 6-100=-94 8-75=-67 10-60=-50 12-50=-38 15-40=-25 20-30=-10 24-25=-1$$

The solution is the pair that gives sum -10.

$$a=-30 b=20$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-30\right)\left(x+20\right)$$

To find equation solutions, solve x-30=0 and x+20=0.

$$x=30 x=-20$$

Steps Using Factoring By Grouping

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

$$x^{2}-20x+100=10\left(70-x\right)$$

Use the distributive property to multiply 10 by 70-x.

$$x^{2}-20x+100=700-10x$$

Subtract 700 from both sides.

$$x^{2}-20x+100-700=-10x$$

Subtract 700 from 100 to get -600.

$$x^{2}-20x-600=-10x$$

Add 10x to both sides.

$$x^{2}-20x-600+10x=0$$

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

$$x^{2}-10x-600=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-600. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-600\right)=-600$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -600.

$$1,-600 2,-300 3,-200 4,-150 5,-120 6,-100 8,-75 10,-60 12,-50 15,-40 20,-30 24,-25$$

Calculate the sum for each pair.

$$1-600=-599 2-300=-298 3-200=-197 4-150=-146 5-120=-115 6-100=-94 8-75=-67 10-60=-50 12-50=-38 15-40=-25 20-30=-10 24-25=-1$$

The solution is the pair that gives sum -10.

$$a=-30 b=20$$

Rewrite x^{2}-10x-600 as \left(x^{2}-30x\right)+\left(20x-600\right).

$$\left(x^{2}-30x\right)+\left(20x-600\right)$$

Factor out x in the first and 20 in the second group.

$$x\left(x-30\right)+20\left(x-30\right)$$

Factor out common term x-30 by using distributive property.

$$\left(x-30\right)\left(x+20\right)$$

To find equation solutions, solve x-30=0 and x+20=0.

$$x=30 x=-20$$

Steps Using the Quadratic Formula

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

$$x^{2}-20x+100=10\left(70-x\right)$$

Use the distributive property to multiply 10 by 70-x.

$$x^{2}-20x+100=700-10x$$

Subtract 700 from both sides.

$$x^{2}-20x+100-700=-10x$$

Subtract 700 from 100 to get -600.

$$x^{2}-20x-600=-10x$$

Add 10x to both sides.

$$x^{2}-20x-600+10x=0$$

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

$$x^{2}-10x-600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-600\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-600\right)}}{2}$$

Multiply -4 times -600.

$$x=\frac{-\left(-10\right)±\sqrt{100+2400}}{2}$$

Add 100 to 2400.

$$x=\frac{-\left(-10\right)±\sqrt{2500}}{2}$$

Take the square root of 2500.

$$x=\frac{-\left(-10\right)±50}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±50}{2}$$

Now solve the equation x=\frac{10±50}{2} when ± is plus. Add 10 to 50.

$$x=\frac{60}{2}$$

Divide 60 by 2.

$$x=30$$

Now solve the equation x=\frac{10±50}{2} when ± is minus. Subtract 50 from 10.

$$x=-\frac{40}{2}$$

Divide -40 by 2.

$$x=-20$$

The equation is now solved.

$$x=30 x=-20$$

Steps for Completing the Square

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

$$x^{2}-20x+100=10\left(70-x\right)$$

Use the distributive property to multiply 10 by 70-x.

$$x^{2}-20x+100=700-10x$$

Add 10x to both sides.

$$x^{2}-20x+100+10x=700$$

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

$$x^{2}-10x+100=700$$

Subtract 100 from both sides.

$$x^{2}-10x=700-100$$

Subtract 100 from 700 to get 600.

$$x^{2}-10x=600$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=600+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=600+25$$

Add 600 to 25.

$$x^{2}-10x+25=625$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=625$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{625}$$

Simplify.

$$x-5=25 x-5=-25$$

Add 5 to both sides of the equation.

$$x=30 x=-20$$