Solve (200-4x)*(100-3x)+(4x*3x)=7700

$$(200-4x) \times (100-3x)+(4x \times 3x)=7700$$

Solve for x (complex solution)

$$x=\frac{125+5\sqrt{113}i}{6}\approx 20.833333333+8.858454844i$$

$$x=\frac{-5\sqrt{113}i+125}{6}\approx 20.833333333-8.858454844i$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$\left(200-4x\right)\left(100-3x\right)+4x^{2}\times 3=7700$$

Use the distributive property to multiply 200-4x by 100-3x and combine like terms.

$$20000-1000x+12x^{2}+4x^{2}\times 3=7700$$

Multiply 4 and 3 to get 12.

$$20000-1000x+12x^{2}+12x^{2}=7700$$

Combine 12x^{2} and 12x^{2} to get 24x^{2}.

$$20000-1000x+24x^{2}=7700$$

Subtract 7700 from both sides.

$$20000-1000x+24x^{2}-7700=0$$

Subtract 7700 from 20000 to get 12300.

$$12300-1000x+24x^{2}=0$$

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.

$$24x^{2}-1000x+12300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, -1000 for b, and 12300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1000\right)±\sqrt{\left(-1000\right)^{2}-4\times 24\times 12300}}{2\times 24}$$

Square -1000.

$$x=\frac{-\left(-1000\right)±\sqrt{1000000-4\times 24\times 12300}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-\left(-1000\right)±\sqrt{1000000-96\times 12300}}{2\times 24}$$

Multiply -96 times 12300.

$$x=\frac{-\left(-1000\right)±\sqrt{1000000-1180800}}{2\times 24}$$

Add 1000000 to -1180800.

$$x=\frac{-\left(-1000\right)±\sqrt{-180800}}{2\times 24}$$

Take the square root of -180800.

$$x=\frac{-\left(-1000\right)±40\sqrt{113}i}{2\times 24}$$

The opposite of -1000 is 1000.

$$x=\frac{1000±40\sqrt{113}i}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{1000±40\sqrt{113}i}{48}$$

Now solve the equation x=\frac{1000±40\sqrt{113}i}{48} when ± is plus. Add 1000 to 40i\sqrt{113}.

$$x=\frac{1000+40\sqrt{113}i}{48}$$

Divide 1000+40i\sqrt{113} by 48.

$$x=\frac{125+5\sqrt{113}i}{6}$$

Now solve the equation x=\frac{1000±40\sqrt{113}i}{48} when ± is minus. Subtract 40i\sqrt{113} from 1000.

$$x=\frac{-40\sqrt{113}i+1000}{48}$$

Divide 1000-40i\sqrt{113} by 48.

$$x=\frac{-5\sqrt{113}i+125}{6}$$

The equation is now solved.

$$x=\frac{125+5\sqrt{113}i}{6} x=\frac{-5\sqrt{113}i+125}{6}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$\left(200-4x\right)\left(100-3x\right)+4x^{2}\times 3=7700$$

Use the distributive property to multiply 200-4x by 100-3x and combine like terms.

$$20000-1000x+12x^{2}+4x^{2}\times 3=7700$$

Multiply 4 and 3 to get 12.

$$20000-1000x+12x^{2}+12x^{2}=7700$$

Combine 12x^{2} and 12x^{2} to get 24x^{2}.

$$20000-1000x+24x^{2}=7700$$

Subtract 20000 from both sides.

$$-1000x+24x^{2}=7700-20000$$

Subtract 20000 from 7700 to get -12300.

$$-1000x+24x^{2}=-12300$$

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.

$$24x^{2}-1000x=-12300$$

Divide both sides by 24.

$$\frac{24x^{2}-1000x}{24}=-\frac{12300}{24}$$

Dividing by 24 undoes the multiplication by 24.

$$x^{2}+\left(-\frac{1000}{24}\right)x=-\frac{12300}{24}$$

Reduce the fraction \frac{-1000}{24} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{125}{3}x=-\frac{12300}{24}$$

Reduce the fraction \frac{-12300}{24} to lowest terms by extracting and canceling out 12.

$$x^{2}-\frac{125}{3}x=-\frac{1025}{2}$$

Divide -\frac{125}{3}, the coefficient of the x term, by 2 to get -\frac{125}{6}. Then add the square of -\frac{125}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{125}{3}x+\left(-\frac{125}{6}\right)^{2}=-\frac{1025}{2}+\left(-\frac{125}{6}\right)^{2}$$

Square -\frac{125}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{125}{3}x+\frac{15625}{36}=-\frac{1025}{2}+\frac{15625}{36}$$

Add -\frac{1025}{2} to \frac{15625}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{125}{3}x+\frac{15625}{36}=-\frac{2825}{36}$$

Factor x^{2}-\frac{125}{3}x+\frac{15625}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{125}{6}\right)^{2}=-\frac{2825}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{125}{6}\right)^{2}}=\sqrt{-\frac{2825}{36}}$$

Simplify.

$$x-\frac{125}{6}=\frac{5\sqrt{113}i}{6} x-\frac{125}{6}=-\frac{5\sqrt{113}i}{6}$$

Add \frac{125}{6} to both sides of the equation.

$$x=\frac{125+5\sqrt{113}i}{6} x=\frac{-5\sqrt{113}i+125}{6}$$