Solve x^2-5x-2250=0

$${ x }^{ 2 } -5x-2250=0$$

Solve for x

$$x=-45$$

$$x=50$$

Steps Using Factoring

To solve the equation, factor x^{2}-5x-2250 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.

$$a+b=-5 ab=-2250$$

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 -2250.

$$1,-2250 2,-1125 3,-750 5,-450 6,-375 9,-250 10,-225 15,-150 18,-125 25,-90 30,-75 45,-50$$

Calculate the sum for each pair.

$$1-2250=-2249 2-1125=-1123 3-750=-747 5-450=-445 6-375=-369 9-250=-241 10-225=-215 15-150=-135 18-125=-107 25-90=-65 30-75=-45 45-50=-5$$

The solution is the pair that gives sum -5.

$$a=-50 b=45$$

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

$$\left(x-50\right)\left(x+45\right)$$

To find equation solutions, solve x-50=0 and x+45=0.

$$x=50 x=-45$$

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-2250. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-2250\right)=-2250$$

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 -2250.

$$1,-2250 2,-1125 3,-750 5,-450 6,-375 9,-250 10,-225 15,-150 18,-125 25,-90 30,-75 45,-50$$

Calculate the sum for each pair.

$$1-2250=-2249 2-1125=-1123 3-750=-747 5-450=-445 6-375=-369 9-250=-241 10-225=-215 15-150=-135 18-125=-107 25-90=-65 30-75=-45 45-50=-5$$

The solution is the pair that gives sum -5.

$$a=-50 b=45$$

Rewrite x^{2}-5x-2250 as \left(x^{2}-50x\right)+\left(45x-2250\right).

$$\left(x^{2}-50x\right)+\left(45x-2250\right)$$

Factor out x in the first and 45 in the second group.

$$x\left(x-50\right)+45\left(x-50\right)$$

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

$$\left(x-50\right)\left(x+45\right)$$

To find equation solutions, solve x-50=0 and x+45=0.

$$x=50 x=-45$$

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.

$$x^{2}-5x-2250=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -2250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2250\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2250\right)}}{2}$$

Multiply -4 times -2250.

$$x=\frac{-\left(-5\right)±\sqrt{25+9000}}{2}$$

Add 25 to 9000.

$$x=\frac{-\left(-5\right)±\sqrt{9025}}{2}$$

Take the square root of 9025.

$$x=\frac{-\left(-5\right)±95}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±95}{2}$$

Now solve the equation x=\frac{5±95}{2} when ± is plus. Add 5 to 95.

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

Divide 100 by 2.

$$x=50$$

Now solve the equation x=\frac{5±95}{2} when ± is minus. Subtract 95 from 5.

$$x=-\frac{90}{2}$$

Divide -90 by 2.

$$x=-45$$

The equation is now solved.

$$x=50 x=-45$$

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.

$$x^{2}-5x-2250=0$$

Add 2250 to both sides of the equation.

$$x^{2}-5x-2250-\left(-2250\right)=-\left(-2250\right)$$

Subtracting -2250 from itself leaves 0.

$$x^{2}-5x=-\left(-2250\right)$$

Subtract -2250 from 0.

$$x^{2}-5x=2250$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=2250+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=2250+\frac{25}{4}$$

Add 2250 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9025}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9025}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9025}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{95}{2} x-\frac{5}{2}=-\frac{95}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=50 x=-45$$