Solve x^2-10x-600=0

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

Solve for x

$$x=-20$$

$$x=30$$

Steps Using Factoring

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

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

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.

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

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Add 600 to both sides of the equation.

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

Subtracting -600 from itself leaves 0.

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

Subtract -600 from 0.

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

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.

$$x=30 x=-20$$