Solve x^2=2000+10x

$$x ^ { 2 } = 2000 + 10 x$$

Solve for x

$$x=-40$$

$$x=50$$

Steps Using Factoring

Subtract 2000 from both sides.

$$x^{2}-2000=10x$$

Subtract 10x from both sides.

$$x^{2}-2000-10x=0$$

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

$$x^{2}-10x-2000=0$$

To solve the equation, factor x^{2}-10x-2000 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.

$$a+b=-10 ab=-2000$$

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 -2000.

$$1,-2000 2,-1000 4,-500 5,-400 8,-250 10,-200 16,-125 20,-100 25,-80 40,-50$$

Calculate the sum for each pair.

$$1-2000=-1999 2-1000=-998 4-500=-496 5-400=-395 8-250=-242 10-200=-190 16-125=-109 20-100=-80 25-80=-55 40-50=-10$$

The solution is the pair that gives sum -10.

$$a=-50 b=40$$

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

$$\left(x-50\right)\left(x+40\right)$$

To find equation solutions, solve x-50=0 and x+40=0.

$$x=50 x=-40$$

Steps Using Factoring By Grouping

Subtract 2000 from both sides.

$$x^{2}-2000=10x$$

Subtract 10x from both sides.

$$x^{2}-2000-10x=0$$

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

$$x^{2}-10x-2000=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-2000. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-2000\right)=-2000$$

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 -2000.

$$1,-2000 2,-1000 4,-500 5,-400 8,-250 10,-200 16,-125 20,-100 25,-80 40,-50$$

Calculate the sum for each pair.

$$1-2000=-1999 2-1000=-998 4-500=-496 5-400=-395 8-250=-242 10-200=-190 16-125=-109 20-100=-80 25-80=-55 40-50=-10$$

The solution is the pair that gives sum -10.

$$a=-50 b=40$$

Rewrite x^{2}-10x-2000 as \left(x^{2}-50x\right)+\left(40x-2000\right).

$$\left(x^{2}-50x\right)+\left(40x-2000\right)$$

Factor out x in the first and 40 in the second group.

$$x\left(x-50\right)+40\left(x-50\right)$$

Factor out common term x-50 by using distributive property.

$$\left(x-50\right)\left(x+40\right)$$

To find equation solutions, solve x-50=0 and x+40=0.

$$x=50 x=-40$$

Steps Using the Quadratic Formula

Subtract 2000 from both sides.

$$x^{2}-2000=10x$$

Subtract 10x from both sides.

$$x^{2}-2000-10x=0$$

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.

$$x^{2}-10x-2000=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -2000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-2000\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-2000\right)}}{2}$$

Multiply -4 times -2000.

$$x=\frac{-\left(-10\right)±\sqrt{100+8000}}{2}$$

Add 100 to 8000.

$$x=\frac{-\left(-10\right)±\sqrt{8100}}{2}$$

Take the square root of 8100.

$$x=\frac{-\left(-10\right)±90}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±90}{2}$$

Now solve the equation x=\frac{10±90}{2} when ± is plus. Add 10 to 90.

$$x=\frac{100}{2}$$

Divide 100 by 2.

$$x=50$$

Now solve the equation x=\frac{10±90}{2} when ± is minus. Subtract 90 from 10.

$$x=-\frac{80}{2}$$

Divide -80 by 2.

$$x=-40$$

The equation is now solved.

$$x=50 x=-40$$

Steps for Completing the Square

Subtract 10x from both sides.

$$x^{2}-10x=2000$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=2000+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=2000+25$$

Add 2000 to 25.

$$x^{2}-10x+25=2025$$

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}.

$$\left(x-5\right)^{2}=2025$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{2025}$$

Simplify.

$$x-5=45 x-5=-45$$

Add 5 to both sides of the equation.

$$x=50 x=-40$$