Solve (1.45*10^-5)=(5+2x)(5+2x)/2^3*1

$$(1.45 \times { 10 }^{ -5 } )= \frac{ (5+2x)(5+2x) }{ { 2 }^{ 3 } \times 1 }$$

Solve for x

$$x=\frac{\sqrt{29}}{1000}-2.5\approx -2.494614835$$

$$x=-\frac{\sqrt{29}}{1000}-2.5\approx -2.505385165$$

Steps Using the Quadratic Formula

Multiply 5+2x and 5+2x to get \left(5+2x\right)^{2}.

$$1.45\times 10^{-5}=\frac{\left(5+2x\right)^{2}}{2^{3}\times 1}$$

Calculate 10 to the power of -5 and get \frac{1}{100000}.

$$1.45\times \frac{1}{100000}=\frac{\left(5+2x\right)^{2}}{2^{3}\times 1}$$

Multiply 1.45 and \frac{1}{100000} to get \frac{29}{2000000}.

$$\frac{29}{2000000}=\frac{\left(5+2x\right)^{2}}{2^{3}\times 1}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+2x\right)^{2}.

$$\frac{29}{2000000}=\frac{25+20x+4x^{2}}{2^{3}\times 1}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{29}{2000000}=\frac{25+20x+4x^{2}}{8\times 1}$$

Multiply 8 and 1 to get 8.

$$\frac{29}{2000000}=\frac{25+20x+4x^{2}}{8}$$

Divide each term of 25+20x+4x^{2} by 8 to get \frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}.

$$\frac{29}{2000000}=\frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}=\frac{29}{2000000}$$

Subtract \frac{29}{2000000} from both sides.

$$\frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}-\frac{29}{2000000}=0$$

Subtract \frac{29}{2000000} from \frac{25}{8} to get \frac{6249971}{2000000}.

$$\frac{6249971}{2000000}+\frac{5}{2}x+\frac{1}{2}x^{2}=0$$

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.

$$\frac{1}{2}x^{2}+\frac{5}{2}x+\frac{6249971}{2000000}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, \frac{5}{2} for b, and \frac{6249971}{2000000} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{5}{2}±\sqrt{\left(\frac{5}{2}\right)^{2}-4\times \frac{1}{2}\times \frac{6249971}{2000000}}}{2\times \frac{1}{2}}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-4\times \frac{1}{2}\times \frac{6249971}{2000000}}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-2\times \frac{6249971}{2000000}}}{2\times \frac{1}{2}}$$

Multiply -2 times \frac{6249971}{2000000}.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-\frac{6249971}{1000000}}}{2\times \frac{1}{2}}$$

Add \frac{25}{4} to -\frac{6249971}{1000000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{29}{1000000}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{29}{1000000}.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{29}}{1000}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{29}}{1000}}{1}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{\sqrt{29}}{1000}}{1} when ± is plus. Add -\frac{5}{2} to \frac{\sqrt{29}}{1000}.

$$x=\frac{\frac{\sqrt{29}}{1000}-\frac{5}{2}}{1}$$

Divide -\frac{5}{2}+\frac{\sqrt{29}}{1000} by 1.

$$x=\frac{\sqrt{29}}{1000}-\frac{5}{2}$$

Now solve the equation x=\frac{-\frac{5}{2}±\frac{\sqrt{29}}{1000}}{1} when ± is minus. Subtract \frac{\sqrt{29}}{1000} from -\frac{5}{2}.

$$x=\frac{-\frac{\sqrt{29}}{1000}-\frac{5}{2}}{1}$$

Divide -\frac{5}{2}-\frac{\sqrt{29}}{1000} by 1.

$$x=-\frac{\sqrt{29}}{1000}-\frac{5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{29}}{1000}-\frac{5}{2} x=-\frac{\sqrt{29}}{1000}-\frac{5}{2}$$

Steps for Completing the Square

Multiply 5+2x and 5+2x to get \left(5+2x\right)^{2}.

$$1.45\times 10^{-5}=\frac{\left(5+2x\right)^{2}}{2^{3}\times 1}$$

Calculate 10 to the power of -5 and get \frac{1}{100000}.

$$1.45\times \frac{1}{100000}=\frac{\left(5+2x\right)^{2}}{2^{3}\times 1}$$

Multiply 1.45 and \frac{1}{100000} to get \frac{29}{2000000}.

$$\frac{29}{2000000}=\frac{\left(5+2x\right)^{2}}{2^{3}\times 1}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+2x\right)^{2}.

$$\frac{29}{2000000}=\frac{25+20x+4x^{2}}{2^{3}\times 1}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{29}{2000000}=\frac{25+20x+4x^{2}}{8\times 1}$$

Multiply 8 and 1 to get 8.

$$\frac{29}{2000000}=\frac{25+20x+4x^{2}}{8}$$

Divide each term of 25+20x+4x^{2} by 8 to get \frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}.

$$\frac{29}{2000000}=\frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{25}{8}+\frac{5}{2}x+\frac{1}{2}x^{2}=\frac{29}{2000000}$$

Subtract \frac{25}{8} from both sides.

$$\frac{5}{2}x+\frac{1}{2}x^{2}=\frac{29}{2000000}-\frac{25}{8}$$

Subtract \frac{25}{8} from \frac{29}{2000000} to get -\frac{6249971}{2000000}.

$$\frac{5}{2}x+\frac{1}{2}x^{2}=-\frac{6249971}{2000000}$$

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.

$$\frac{1}{2}x^{2}+\frac{5}{2}x=-\frac{6249971}{2000000}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+\frac{5}{2}x}{\frac{1}{2}}=-\frac{\frac{6249971}{2000000}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{\frac{5}{2}}{\frac{1}{2}}x=-\frac{\frac{6249971}{2000000}}{\frac{1}{2}}$$

Divide \frac{5}{2} by \frac{1}{2} by multiplying \frac{5}{2} by the reciprocal of \frac{1}{2}.

$$x^{2}+5x=-\frac{\frac{6249971}{2000000}}{\frac{1}{2}}$$

Divide -\frac{6249971}{2000000} by \frac{1}{2} by multiplying -\frac{6249971}{2000000} by the reciprocal of \frac{1}{2}.

$$x^{2}+5x=-\frac{6249971}{1000000}$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{6249971}{1000000}+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-\frac{6249971}{1000000}+\frac{25}{4}$$

Add -\frac{6249971}{1000000} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=\frac{29}{1000000}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{29}{1000000}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{29}{1000000}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{29}}{1000} x+\frac{5}{2}=-\frac{\sqrt{29}}{1000}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{29}}{1000}-\frac{5}{2} x=-\frac{\sqrt{29}}{1000}-\frac{5}{2}$$