Solve 3000=(125+x)(45-x)

$$3000=(125+x)(45-x)$$

Solve for x

$$x=-105$$

$$x=25$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 125+x by 45-x and combine like terms.

$$3000=5625-80x-x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5625-80x-x^{2}=3000$$

Subtract 3000 from both sides.

$$5625-80x-x^{2}-3000=0$$

Subtract 3000 from 5625 to get 2625.

$$2625-80x-x^{2}=0$$

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.

$$-x^{2}-80x+2625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -80 for b, and 2625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\left(-1\right)\times 2625}}{2\left(-1\right)}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\left(-1\right)\times 2625}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-80\right)±\sqrt{6400+4\times 2625}}{2\left(-1\right)}$$

Multiply 4 times 2625.

$$x=\frac{-\left(-80\right)±\sqrt{6400+10500}}{2\left(-1\right)}$$

Add 6400 to 10500.

$$x=\frac{-\left(-80\right)±\sqrt{16900}}{2\left(-1\right)}$$

Take the square root of 16900.

$$x=\frac{-\left(-80\right)±130}{2\left(-1\right)}$$

The opposite of -80 is 80.

$$x=\frac{80±130}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{80±130}{-2}$$

Now solve the equation x=\frac{80±130}{-2} when ± is plus. Add 80 to 130.

$$x=\frac{210}{-2}$$

Divide 210 by -2.

$$x=-105$$

Now solve the equation x=\frac{80±130}{-2} when ± is minus. Subtract 130 from 80.

$$x=-\frac{50}{-2}$$

Divide -50 by -2.

$$x=25$$

The equation is now solved.

$$x=-105 x=25$$

Steps for Completing the Square

Use the distributive property to multiply 125+x by 45-x and combine like terms.

$$3000=5625-80x-x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5625-80x-x^{2}=3000$$

Subtract 5625 from both sides.

$$-80x-x^{2}=3000-5625$$

Subtract 5625 from 3000 to get -2625.

$$-80x-x^{2}=-2625$$

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.

$$-x^{2}-80x=-2625$$

Divide both sides by -1.

$$\frac{-x^{2}-80x}{-1}=-\frac{2625}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{80}{-1}\right)x=-\frac{2625}{-1}$$

Divide -80 by -1.

$$x^{2}+80x=-\frac{2625}{-1}$$

Divide -2625 by -1.

$$x^{2}+80x=2625$$

Divide 80, the coefficient of the x term, by 2 to get 40. Then add the square of 40 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+80x+40^{2}=2625+40^{2}$$

Square 40.

$$x^{2}+80x+1600=2625+1600$$

Add 2625 to 1600.

$$x^{2}+80x+1600=4225$$

Factor x^{2}+80x+1600. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+40\right)^{2}=4225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+40\right)^{2}}=\sqrt{4225}$$

Simplify.

$$x+40=65 x+40=-65$$

Subtract 40 from both sides of the equation.

$$x=25 x=-105$$