Solve 3=9-20x/9-400x^2

$$3 = \frac { 9 - 20 x } { 9 - 400 x ^ { 2 } }$$

Solve for x

$$x=\frac{\sqrt{217}+1}{120}\approx 0.131090999$$

$$x=\frac{1-\sqrt{217}}{120}\approx -0.114424332$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{3}{20},\frac{3}{20} since division by zero is not defined. Multiply both sides of the equation by \left(-20x-3\right)\left(20x-3\right).

$$3\left(-20x-3\right)\left(20x-3\right)=9-20x$$

Use the distributive property to multiply 3 by -20x-3.

$$\left(-60x-9\right)\left(20x-3\right)=9-20x$$

Use the distributive property to multiply -60x-9 by 20x-3 and combine like terms.

$$-1200x^{2}+27=9-20x$$

Subtract 9 from both sides.

$$-1200x^{2}+27-9=-20x$$

Subtract 9 from 27 to get 18.

$$-1200x^{2}+18=-20x$$

Add 20x to both sides.

$$-1200x^{2}+18+20x=0$$

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.

$$-1200x^{2}+20x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1200 for a, 20 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-1200\right)\times 18}}{2\left(-1200\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-1200\right)\times 18}}{2\left(-1200\right)}$$

Multiply -4 times -1200.

$$x=\frac{-20±\sqrt{400+4800\times 18}}{2\left(-1200\right)}$$

Multiply 4800 times 18.

$$x=\frac{-20±\sqrt{400+86400}}{2\left(-1200\right)}$$

Add 400 to 86400.

$$x=\frac{-20±\sqrt{86800}}{2\left(-1200\right)}$$

Take the square root of 86800.

$$x=\frac{-20±20\sqrt{217}}{2\left(-1200\right)}$$

Multiply 2 times -1200.

$$x=\frac{-20±20\sqrt{217}}{-2400}$$

Now solve the equation x=\frac{-20±20\sqrt{217}}{-2400} when ± is plus. Add -20 to 20\sqrt{217}.

$$x=\frac{20\sqrt{217}-20}{-2400}$$

Divide -20+20\sqrt{217} by -2400.

$$x=\frac{1-\sqrt{217}}{120}$$

Now solve the equation x=\frac{-20±20\sqrt{217}}{-2400} when ± is minus. Subtract 20\sqrt{217} from -20.

$$x=\frac{-20\sqrt{217}-20}{-2400}$$

Divide -20-20\sqrt{217} by -2400.

$$x=\frac{\sqrt{217}+1}{120}$$

The equation is now solved.

$$x=\frac{1-\sqrt{217}}{120} x=\frac{\sqrt{217}+1}{120}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{3}{20},\frac{3}{20} since division by zero is not defined. Multiply both sides of the equation by \left(-20x-3\right)\left(20x-3\right).

$$3\left(-20x-3\right)\left(20x-3\right)=9-20x$$

Use the distributive property to multiply 3 by -20x-3.

$$\left(-60x-9\right)\left(20x-3\right)=9-20x$$

Use the distributive property to multiply -60x-9 by 20x-3 and combine like terms.

$$-1200x^{2}+27=9-20x$$

Add 20x to both sides.

$$-1200x^{2}+27+20x=9$$

Subtract 27 from both sides.

$$-1200x^{2}+20x=9-27$$

Subtract 27 from 9 to get -18.

$$-1200x^{2}+20x=-18$$

Divide both sides by -1200.

$$\frac{-1200x^{2}+20x}{-1200}=-\frac{18}{-1200}$$

Dividing by -1200 undoes the multiplication by -1200.

$$x^{2}+\frac{20}{-1200}x=-\frac{18}{-1200}$$

Reduce the fraction \frac{20}{-1200} to lowest terms by extracting and canceling out 20.

$$x^{2}-\frac{1}{60}x=-\frac{18}{-1200}$$

Reduce the fraction \frac{-18}{-1200} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{1}{60}x=\frac{3}{200}$$

Divide -\frac{1}{60}, the coefficient of the x term, by 2 to get -\frac{1}{120}. Then add the square of -\frac{1}{120} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{60}x+\left(-\frac{1}{120}\right)^{2}=\frac{3}{200}+\left(-\frac{1}{120}\right)^{2}$$

Square -\frac{1}{120} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{60}x+\frac{1}{14400}=\frac{3}{200}+\frac{1}{14400}$$

Add \frac{3}{200} to \frac{1}{14400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{60}x+\frac{1}{14400}=\frac{217}{14400}$$

Factor x^{2}-\frac{1}{60}x+\frac{1}{14400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{120}\right)^{2}=\frac{217}{14400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{120}\right)^{2}}=\sqrt{\frac{217}{14400}}$$

Simplify.

$$x-\frac{1}{120}=\frac{\sqrt{217}}{120} x-\frac{1}{120}=-\frac{\sqrt{217}}{120}$$

Add \frac{1}{120} to both sides of the equation.

$$x=\frac{\sqrt{217}+1}{120} x=\frac{1-\sqrt{217}}{120}$$