Solve 25x^2-4=0

$$25 x ^ { 2 } - 4 = 0$$

Solve for x

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

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

Steps Using Difference of Squares

Consider 25x^{2}-4. Rewrite 25x^{2}-4 as \left(5x\right)^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(5x-2\right)\left(5x+2\right)=0$$

To find equation solutions, solve 5x-2=0 and 5x+2=0.

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

Steps by Finding Square Root

Add 4 to both sides. Anything plus zero gives itself.

$$25x^{2}=4$$

Divide both sides by 25.

$$x^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

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

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.

$$25x^{2}-4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\times 25\left(-4\right)}}{2\times 25}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 25\left(-4\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{0±\sqrt{-100\left(-4\right)}}{2\times 25}$$

Multiply -100 times -4.

$$x=\frac{0±\sqrt{400}}{2\times 25}$$

Take the square root of 400.

$$x=\frac{0±20}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{0±20}{50}$$

Now solve the equation x=\frac{0±20}{50} when ± is plus. Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{0±20}{50} when ± is minus. Reduce the fraction \frac{-20}{50} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

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