Solve (40-2x)(25-2x)=450

$$(40-2x)(25-2x)=450$$

Solve for x

$$x = \frac{55}{2} = 27\frac{1}{2} = 27.5$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 40-2x by 25-2x and combine like terms.

$$1000-130x+4x^{2}=450$$

Subtract 450 from both sides.

$$1000-130x+4x^{2}-450=0$$

Subtract 450 from 1000 to get 550.

$$550-130x+4x^{2}=0$$

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.

$$4x^{2}-130x+550=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -130 for b, and 550 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-130\right)±\sqrt{\left(-130\right)^{2}-4\times 4\times 550}}{2\times 4}$$

Square -130.

$$x=\frac{-\left(-130\right)±\sqrt{16900-4\times 4\times 550}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-130\right)±\sqrt{16900-16\times 550}}{2\times 4}$$

Multiply -16 times 550.

$$x=\frac{-\left(-130\right)±\sqrt{16900-8800}}{2\times 4}$$

Add 16900 to -8800.

$$x=\frac{-\left(-130\right)±\sqrt{8100}}{2\times 4}$$

Take the square root of 8100.

$$x=\frac{-\left(-130\right)±90}{2\times 4}$$

The opposite of -130 is 130.

$$x=\frac{130±90}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{130±90}{8}$$

Now solve the equation x=\frac{130±90}{8} when ± is plus. Add 130 to 90.

$$x=\frac{220}{8}$$

Reduce the fraction \frac{220}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{55}{2}$$

Now solve the equation x=\frac{130±90}{8} when ± is minus. Subtract 90 from 130.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

The equation is now solved.

$$x=\frac{55}{2} x=5$$

Steps for Completing the Square

Use the distributive property to multiply 40-2x by 25-2x and combine like terms.

$$1000-130x+4x^{2}=450$$

Subtract 1000 from both sides.

$$-130x+4x^{2}=450-1000$$

Subtract 1000 from 450 to get -550.

$$-130x+4x^{2}=-550$$

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.

$$4x^{2}-130x=-550$$

Divide both sides by 4.

$$\frac{4x^{2}-130x}{4}=-\frac{550}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{130}{4}\right)x=-\frac{550}{4}$$

Reduce the fraction \frac{-130}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{65}{2}x=-\frac{550}{4}$$

Reduce the fraction \frac{-550}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{65}{2}x=-\frac{275}{2}$$

Divide -\frac{65}{2}, the coefficient of the x term, by 2 to get -\frac{65}{4}. Then add the square of -\frac{65}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{65}{2}x+\left(-\frac{65}{4}\right)^{2}=-\frac{275}{2}+\left(-\frac{65}{4}\right)^{2}$$

Square -\frac{65}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{65}{2}x+\frac{4225}{16}=-\frac{275}{2}+\frac{4225}{16}$$

Add -\frac{275}{2} to \frac{4225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{65}{2}x+\frac{4225}{16}=\frac{2025}{16}$$

Factor x^{2}-\frac{65}{2}x+\frac{4225}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{65}{4}\right)^{2}=\frac{2025}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{65}{4}\right)^{2}}=\sqrt{\frac{2025}{16}}$$

Simplify.

$$x-\frac{65}{4}=\frac{45}{4} x-\frac{65}{4}=-\frac{45}{4}$$

Add \frac{65}{4} to both sides of the equation.

$$x=\frac{55}{2} x=5$$