Solve 136*10^-2=x^2/(0390-x)

$$136 \times 10 ^ { - 2 } = \frac { x ^ { 2 } } { ( 0390 - x ) }$$

Solve for x

$$x = -\frac{34}{25} = -1\frac{9}{25} = -1.36$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$136\times 10^{-2}x=-x^{2}$$

Calculate 10 to the power of -2 and get \frac{1}{100}.

$$136\times \frac{1}{100}x=-x^{2}$$

Multiply 136 and \frac{1}{100} to get \frac{34}{25}.

$$\frac{34}{25}x=-x^{2}$$

Add x^{2} to both sides.

$$\frac{34}{25}x+x^{2}=0$$

Factor out x.

$$x\left(\frac{34}{25}+x\right)=0$$

To find equation solutions, solve x=0 and \frac{34}{25}+x=0.

$$x=0 x=-\frac{34}{25}$$

Variable x cannot be equal to 0.

$$x=-\frac{34}{25}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$136\times 10^{-2}x=-x^{2}$$

Calculate 10 to the power of -2 and get \frac{1}{100}.

$$136\times \frac{1}{100}x=-x^{2}$$

Multiply 136 and \frac{1}{100} to get \frac{34}{25}.

$$\frac{34}{25}x=-x^{2}$$

Add x^{2} to both sides.

$$\frac{34}{25}x+x^{2}=0$$

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.

$$x^{2}+\frac{34}{25}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, \frac{34}{25} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{34}{25}±\sqrt{\left(\frac{34}{25}\right)^{2}}}{2}$$

Take the square root of \left(\frac{34}{25}\right)^{2}.

$$x=\frac{-\frac{34}{25}±\frac{34}{25}}{2}$$

Now solve the equation x=\frac{-\frac{34}{25}±\frac{34}{25}}{2} when ± is plus. Add -\frac{34}{25} to \frac{34}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

Now solve the equation x=\frac{-\frac{34}{25}±\frac{34}{25}}{2} when ± is minus. Subtract \frac{34}{25} from -\frac{34}{25} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{68}{25}}{2}$$

Divide -\frac{68}{25} by 2.

$$x=-\frac{34}{25}$$

The equation is now solved.

$$x=0 x=-\frac{34}{25}$$

Variable x cannot be equal to 0.

$$x=-\frac{34}{25}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$136\times 10^{-2}x=-x^{2}$$

Calculate 10 to the power of -2 and get \frac{1}{100}.

$$136\times \frac{1}{100}x=-x^{2}$$

Multiply 136 and \frac{1}{100} to get \frac{34}{25}.

$$\frac{34}{25}x=-x^{2}$$

Add x^{2} to both sides.

$$\frac{34}{25}x+x^{2}=0$$

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.

$$x^{2}+\frac{34}{25}x=0$$

Divide \frac{34}{25}, the coefficient of the x term, by 2 to get \frac{17}{25}. Then add the square of \frac{17}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{34}{25}x+\left(\frac{17}{25}\right)^{2}=\left(\frac{17}{25}\right)^{2}$$

Square \frac{17}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{34}{25}x+\frac{289}{625}=\frac{289}{625}$$

Factor x^{2}+\frac{34}{25}x+\frac{289}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{17}{25}\right)^{2}=\frac{289}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{25}\right)^{2}}=\sqrt{\frac{289}{625}}$$

Simplify.

$$x+\frac{17}{25}=\frac{17}{25} x+\frac{17}{25}=-\frac{17}{25}$$

Subtract \frac{17}{25} from both sides of the equation.

$$x=0 x=-\frac{34}{25}$$

Variable x cannot be equal to 0.

$$x=-\frac{34}{25}$$