Solve 10x^2-1=3/2x

$$10 x ^ { 2 } - 1 = \frac { 3 } { 2 } x$$

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=\frac{2}{5}=0.4$$

Steps Using the Quadratic Formula

Subtract \frac{3}{2}x from both sides.

$$10x^{2}-1-\frac{3}{2}x=0$$

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.

$$10x^{2}-\frac{3}{2}x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -\frac{3}{2} for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\left(-\frac{3}{2}\right)^{2}-4\times 10\left(-1\right)}}{2\times 10}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4\times 10\left(-1\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-40\left(-1\right)}}{2\times 10}$$

Multiply -40 times -1.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}+40}}{2\times 10}$$

Add \frac{9}{4} to 40.

$$x=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{169}{4}}}{2\times 10}$$

Take the square root of \frac{169}{4}.

$$x=\frac{-\left(-\frac{3}{2}\right)±\frac{13}{2}}{2\times 10}$$

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$x=\frac{\frac{3}{2}±\frac{13}{2}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{\frac{3}{2}±\frac{13}{2}}{20}$$

Now solve the equation x=\frac{\frac{3}{2}±\frac{13}{2}}{20} when ± is plus. Add \frac{3}{2} to \frac{13}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{8}{20}$$

Reduce the fraction \frac{8}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{2}{5}$$

Now solve the equation x=\frac{\frac{3}{2}±\frac{13}{2}}{20} when ± is minus. Subtract \frac{13}{2} from \frac{3}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{5}{20}$$

Reduce the fraction \frac{-5}{20} to lowest terms by extracting and canceling out 5.

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=\frac{2}{5} x=-\frac{1}{4}$$

Steps for Completing the Square

Subtract \frac{3}{2}x from both sides.

$$10x^{2}-1-\frac{3}{2}x=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$10x^{2}-\frac{3}{2}x=1$$

Divide both sides by 10.

$$\frac{10x^{2}-\frac{3}{2}x}{10}=\frac{1}{10}$$

Dividing by 10 undoes the multiplication by 10.

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

Divide -\frac{3}{2} by 10.

$$x^{2}-\frac{3}{20}x=\frac{1}{10}$$

Divide -\frac{3}{20}, the coefficient of the x term, by 2 to get -\frac{3}{40}. Then add the square of -\frac{3}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{20}x+\left(-\frac{3}{40}\right)^{2}=\frac{1}{10}+\left(-\frac{3}{40}\right)^{2}$$

Square -\frac{3}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{20}x+\frac{9}{1600}=\frac{1}{10}+\frac{9}{1600}$$

Add \frac{1}{10} to \frac{9}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{20}x+\frac{9}{1600}=\frac{169}{1600}$$

Factor x^{2}-\frac{3}{20}x+\frac{9}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{40}\right)^{2}=\frac{169}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{40}\right)^{2}}=\sqrt{\frac{169}{1600}}$$

Simplify.

$$x-\frac{3}{40}=\frac{13}{40} x-\frac{3}{40}=-\frac{13}{40}$$

Add \frac{3}{40} to both sides of the equation.

$$x=\frac{2}{5} x=-\frac{1}{4}$$