Solve 6x^2-3/8=0

$$6 x ^ { 2 } - \frac { 3 } { 8 } = 0$$

Solve for x

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

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

Steps Using Difference of Squares

Divide both sides by \frac{3}{8}.

$$16x^{2}-1=0$$

Consider 16x^{2}-1. Rewrite 16x^{2}-1 as \left(4x\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4x-1\right)\left(4x+1\right)=0$$

To find equation solutions, solve 4x-1=0 and 4x+1=0.

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

Steps by Finding Square Root

Add \frac{3}{8} to both sides. Anything plus zero gives itself.

$$6x^{2}=\frac{3}{8}$$

Divide both sides by 6.

$$x^{2}=\frac{\frac{3}{8}}{6}$$

Express \frac{\frac{3}{8}}{6} as a single fraction.

$$x^{2}=\frac{3}{8\times 6}$$

Multiply 8 and 6 to get 48.

$$x^{2}=\frac{3}{48}$$

Reduce the fraction \frac{3}{48} to lowest terms by extracting and canceling out 3.

$$x^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$6x^{2}-\frac{3}{8}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 0 for b, and -\frac{3}{8} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 6\left(-\frac{3}{8}\right)}}{2\times 6}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 6\left(-\frac{3}{8}\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{0±\sqrt{-24\left(-\frac{3}{8}\right)}}{2\times 6}$$

Multiply -24 times -\frac{3}{8}.

$$x=\frac{0±\sqrt{9}}{2\times 6}$$

Take the square root of 9.

$$x=\frac{0±3}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{0±3}{12}$$

Now solve the equation x=\frac{0±3}{12} when ± is plus. Reduce the fraction \frac{3}{12} to lowest terms by extracting and canceling out 3.

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

Now solve the equation x=\frac{0±3}{12} when ± is minus. Reduce the fraction \frac{-3}{12} to lowest terms by extracting and canceling out 3.

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

The equation is now solved.

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