Solve (60-x-40)(20x+300)=6120

$$( 60 - x - 40 ) ( 20 x + 300 ) = 6120$$

Solve for x

$$x=2$$

$$x=3$$

Steps Using the Quadratic Formula

Subtract 40 from 60 to get 20.

$$\left(20-x\right)\left(20x+300\right)=6120$$

Use the distributive property to multiply 20-x by 20x+300 and combine like terms.

$$100x+6000-20x^{2}=6120$$

Subtract 6120 from both sides.

$$100x+6000-20x^{2}-6120=0$$

Subtract 6120 from 6000 to get -120.

$$100x-120-20x^{2}=0$$

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.

$$-20x^{2}+100x-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 100 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-100±\sqrt{100^{2}-4\left(-20\right)\left(-120\right)}}{2\left(-20\right)}$$

Square 100.

$$x=\frac{-100±\sqrt{10000-4\left(-20\right)\left(-120\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-100±\sqrt{10000+80\left(-120\right)}}{2\left(-20\right)}$$

Multiply 80 times -120.

$$x=\frac{-100±\sqrt{10000-9600}}{2\left(-20\right)}$$

Add 10000 to -9600.

$$x=\frac{-100±\sqrt{400}}{2\left(-20\right)}$$

Take the square root of 400.

$$x=\frac{-100±20}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-100±20}{-40}$$

Now solve the equation x=\frac{-100±20}{-40} when ± is plus. Add -100 to 20.

$$x=-\frac{80}{-40}$$

Divide -80 by -40.

$$x=2$$

Now solve the equation x=\frac{-100±20}{-40} when ± is minus. Subtract 20 from -100.

$$x=-\frac{120}{-40}$$

Divide -120 by -40.

$$x=3$$

The equation is now solved.

$$x=2 x=3$$

Steps for Completing the Square

Subtract 40 from 60 to get 20.

$$\left(20-x\right)\left(20x+300\right)=6120$$

Use the distributive property to multiply 20-x by 20x+300 and combine like terms.

$$100x+6000-20x^{2}=6120$$

Subtract 6000 from both sides.

$$100x-20x^{2}=6120-6000$$

Subtract 6000 from 6120 to get 120.

$$100x-20x^{2}=120$$

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.

$$-20x^{2}+100x=120$$

Divide both sides by -20.

$$\frac{-20x^{2}+100x}{-20}=\frac{120}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$x^{2}+\frac{100}{-20}x=\frac{120}{-20}$$

Divide 100 by -20.

$$x^{2}-5x=\frac{120}{-20}$$

Divide 120 by -20.

$$x^{2}-5x=-6$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

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

$$x=3 x=2$$