Solve (50-2x)(60-3x)-x(60-3x)/2=2430

$$( 50 - 2 x ) ( 60 - 3 x ) - \frac { x ( 60 - 3 x ) } { 2 } = 2430$$

Solve for x

$$x=2$$

$$x=38$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2\left(50-2x\right)\left(60-3x\right)-x\left(60-3x\right)=4860$$

Use the distributive property to multiply 2 by 50-2x.

$$\left(100-4x\right)\left(60-3x\right)-x\left(60-3x\right)=4860$$

Use the distributive property to multiply 100-4x by 60-3x and combine like terms.

$$6000-540x+12x^{2}-x\left(60-3x\right)=4860$$

Use the distributive property to multiply x by 60-3x.

$$6000-540x+12x^{2}-\left(60x-3x^{2}\right)=4860$$

To find the opposite of 60x-3x^{2}, find the opposite of each term.

$$6000-540x+12x^{2}-60x+3x^{2}=4860$$

Combine -540x and -60x to get -600x.

$$6000-600x+12x^{2}+3x^{2}=4860$$

Combine 12x^{2} and 3x^{2} to get 15x^{2}.

$$6000-600x+15x^{2}=4860$$

Subtract 4860 from both sides.

$$6000-600x+15x^{2}-4860=0$$

Subtract 4860 from 6000 to get 1140.

$$1140-600x+15x^{2}=0$$

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.

$$15x^{2}-600x+1140=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -600 for b, and 1140 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-600\right)±\sqrt{\left(-600\right)^{2}-4\times 15\times 1140}}{2\times 15}$$

Square -600.

$$x=\frac{-\left(-600\right)±\sqrt{360000-4\times 15\times 1140}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-600\right)±\sqrt{360000-60\times 1140}}{2\times 15}$$

Multiply -60 times 1140.

$$x=\frac{-\left(-600\right)±\sqrt{360000-68400}}{2\times 15}$$

Add 360000 to -68400.

$$x=\frac{-\left(-600\right)±\sqrt{291600}}{2\times 15}$$

Take the square root of 291600.

$$x=\frac{-\left(-600\right)±540}{2\times 15}$$

The opposite of -600 is 600.

$$x=\frac{600±540}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{600±540}{30}$$

Now solve the equation x=\frac{600±540}{30} when ± is plus. Add 600 to 540.

$$x=\frac{1140}{30}$$

Divide 1140 by 30.

$$x=38$$

Now solve the equation x=\frac{600±540}{30} when ± is minus. Subtract 540 from 600.

$$x=\frac{60}{30}$$

Divide 60 by 30.

$$x=2$$

The equation is now solved.

$$x=38 x=2$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2\left(50-2x\right)\left(60-3x\right)-x\left(60-3x\right)=4860$$

Use the distributive property to multiply 2 by 50-2x.

$$\left(100-4x\right)\left(60-3x\right)-x\left(60-3x\right)=4860$$

Use the distributive property to multiply 100-4x by 60-3x and combine like terms.

$$6000-540x+12x^{2}-x\left(60-3x\right)=4860$$

Use the distributive property to multiply x by 60-3x.

$$6000-540x+12x^{2}-\left(60x-3x^{2}\right)=4860$$

To find the opposite of 60x-3x^{2}, find the opposite of each term.

$$6000-540x+12x^{2}-60x+3x^{2}=4860$$

Combine -540x and -60x to get -600x.

$$6000-600x+12x^{2}+3x^{2}=4860$$

Combine 12x^{2} and 3x^{2} to get 15x^{2}.

$$6000-600x+15x^{2}=4860$$

Subtract 6000 from both sides.

$$-600x+15x^{2}=4860-6000$$

Subtract 6000 from 4860 to get -1140.

$$-600x+15x^{2}=-1140$$

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.

$$15x^{2}-600x=-1140$$

Divide both sides by 15.

$$\frac{15x^{2}-600x}{15}=-\frac{1140}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\left(-\frac{600}{15}\right)x=-\frac{1140}{15}$$

Divide -600 by 15.

$$x^{2}-40x=-\frac{1140}{15}$$

Divide -1140 by 15.

$$x^{2}-40x=-76$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=-76+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=-76+400$$

Add -76 to 400.

$$x^{2}-40x+400=324$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=324$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{324}$$

Simplify.

$$x-20=18 x-20=-18$$

Add 20 to both sides of the equation.

$$x=38 x=2$$