Solve 20x*32x+3x=100

$$20x \times 32x+3x=100$$

Solve for x

$$x=\frac{\sqrt{256009}-3}{1280}\approx 0.392947906$$

$$x=\frac{-\sqrt{256009}-3}{1280}\approx -0.397635406$$

Steps Using the Quadratic Formula

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

$$20x^{2}\times 32+3x=100$$

Multiply 20 and 32 to get 640.

$$640x^{2}+3x=100$$

Subtract 100 from both sides.

$$640x^{2}+3x-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 640 for a, 3 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 640\left(-100\right)}}{2\times 640}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 640\left(-100\right)}}{2\times 640}$$

Multiply -4 times 640.

$$x=\frac{-3±\sqrt{9-2560\left(-100\right)}}{2\times 640}$$

Multiply -2560 times -100.

$$x=\frac{-3±\sqrt{9+256000}}{2\times 640}$$

Add 9 to 256000.

$$x=\frac{-3±\sqrt{256009}}{2\times 640}$$

Multiply 2 times 640.

$$x=\frac{-3±\sqrt{256009}}{1280}$$

Now solve the equation x=\frac{-3±\sqrt{256009}}{1280} when ± is plus. Add -3 to \sqrt{256009}.

$$x=\frac{\sqrt{256009}-3}{1280}$$

Now solve the equation x=\frac{-3±\sqrt{256009}}{1280} when ± is minus. Subtract \sqrt{256009} from -3.

$$x=\frac{-\sqrt{256009}-3}{1280}$$

The equation is now solved.

$$x=\frac{\sqrt{256009}-3}{1280} x=\frac{-\sqrt{256009}-3}{1280}$$

Steps for Completing the Square

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

$$20x^{2}\times 32+3x=100$$

Multiply 20 and 32 to get 640.

$$640x^{2}+3x=100$$

Divide both sides by 640.

$$\frac{640x^{2}+3x}{640}=\frac{100}{640}$$

Dividing by 640 undoes the multiplication by 640.

$$x^{2}+\frac{3}{640}x=\frac{100}{640}$$

Reduce the fraction \frac{100}{640} to lowest terms by extracting and canceling out 20.

$$x^{2}+\frac{3}{640}x=\frac{5}{32}$$

Divide \frac{3}{640}, the coefficient of the x term, by 2 to get \frac{3}{1280}. Then add the square of \frac{3}{1280} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{640}x+\left(\frac{3}{1280}\right)^{2}=\frac{5}{32}+\left(\frac{3}{1280}\right)^{2}$$

Square \frac{3}{1280} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{640}x+\frac{9}{1638400}=\frac{5}{32}+\frac{9}{1638400}$$

Add \frac{5}{32} to \frac{9}{1638400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{640}x+\frac{9}{1638400}=\frac{256009}{1638400}$$

Factor x^{2}+\frac{3}{640}x+\frac{9}{1638400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{1280}\right)^{2}=\frac{256009}{1638400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{1280}\right)^{2}}=\sqrt{\frac{256009}{1638400}}$$

Simplify.

$$x+\frac{3}{1280}=\frac{\sqrt{256009}}{1280} x+\frac{3}{1280}=-\frac{\sqrt{256009}}{1280}$$

Subtract \frac{3}{1280} from both sides of the equation.

$$x=\frac{\sqrt{256009}-3}{1280} x=\frac{-\sqrt{256009}-3}{1280}$$