Solve 2000-80x-100x+4x^2=1200

$$2000 - 80 x - 100 x + 4 x ^ { 2 } = 1200$$

Solve for x

$$x=5$$

$$x=40$$

Steps Using the Quadratic Formula

Combine -80x and -100x to get -180x.

$$2000-180x+4x^{2}=1200$$

Subtract 1200 from both sides.

$$2000-180x+4x^{2}-1200=0$$

Subtract 1200 from 2000 to get 800.

$$800-180x+4x^{2}=0$$

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.

$$4x^{2}-180x+800=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -180 for b, and 800 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-180\right)±\sqrt{\left(-180\right)^{2}-4\times 4\times 800}}{2\times 4}$$

Square -180.

$$x=\frac{-\left(-180\right)±\sqrt{32400-4\times 4\times 800}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-180\right)±\sqrt{32400-16\times 800}}{2\times 4}$$

Multiply -16 times 800.

$$x=\frac{-\left(-180\right)±\sqrt{32400-12800}}{2\times 4}$$

Add 32400 to -12800.

$$x=\frac{-\left(-180\right)±\sqrt{19600}}{2\times 4}$$

Take the square root of 19600.

$$x=\frac{-\left(-180\right)±140}{2\times 4}$$

The opposite of -180 is 180.

$$x=\frac{180±140}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{180±140}{8}$$

Now solve the equation x=\frac{180±140}{8} when ± is plus. Add 180 to 140.

$$x=\frac{320}{8}$$

Divide 320 by 8.

$$x=40$$

Now solve the equation x=\frac{180±140}{8} when ± is minus. Subtract 140 from 180.

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

Divide 40 by 8.

$$x=5$$

The equation is now solved.

$$x=40 x=5$$

Steps for Completing the Square

Combine -80x and -100x to get -180x.

$$2000-180x+4x^{2}=1200$$

Subtract 2000 from both sides.

$$-180x+4x^{2}=1200-2000$$

Subtract 2000 from 1200 to get -800.

$$-180x+4x^{2}=-800$$

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.

$$4x^{2}-180x=-800$$

Divide both sides by 4.

$$\frac{4x^{2}-180x}{4}=-\frac{800}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{180}{4}\right)x=-\frac{800}{4}$$

Divide -180 by 4.

$$x^{2}-45x=-\frac{800}{4}$$

Divide -800 by 4.

$$x^{2}-45x=-200$$

Divide -45, the coefficient of the x term, by 2 to get -\frac{45}{2}. Then add the square of -\frac{45}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-45x+\left(-\frac{45}{2}\right)^{2}=-200+\left(-\frac{45}{2}\right)^{2}$$

Square -\frac{45}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-45x+\frac{2025}{4}=-200+\frac{2025}{4}$$

Add -200 to \frac{2025}{4}.

$$x^{2}-45x+\frac{2025}{4}=\frac{1225}{4}$$

Factor x^{2}-45x+\frac{2025}{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}.

$$\left(x-\frac{45}{2}\right)^{2}=\frac{1225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{45}{2}\right)^{2}}=\sqrt{\frac{1225}{4}}$$

Simplify.

$$x-\frac{45}{2}=\frac{35}{2} x-\frac{45}{2}=-\frac{35}{2}$$

Add \frac{45}{2} to both sides of the equation.

$$x=40 x=5$$