Solve 25a^2=49

$$25 a ^ { 2 } = 49$$

Solve for a

$$a = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

$$a = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

Steps Using Difference of Squares

Divide both sides by 25.

$$a^{2}=\frac{49}{25}$$

Subtract \frac{49}{25} from both sides.

$$a^{2}-\frac{49}{25}=0$$

Multiply both sides by 25.

$$25a^{2}-49=0$$

Consider 25a^{2}-49. Rewrite 25a^{2}-49 as \left(5a\right)^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(5a-7\right)\left(5a+7\right)=0$$

To find equation solutions, solve 5a-7=0 and 5a+7=0.

$$a=\frac{7}{5} a=-\frac{7}{5}$$

Steps by Finding Square Root

Divide both sides by 25.

$$a^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$a=\frac{7}{5} a=-\frac{7}{5}$$

Steps Using the Quadratic Formula

Divide both sides by 25.

$$a^{2}=\frac{49}{25}$$

Subtract \frac{49}{25} from both sides.

$$a^{2}-\frac{49}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{49}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-\frac{49}{25}\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-\frac{49}{25}\right)}}{2}$$

Multiply -4 times -\frac{49}{25}.

$$a=\frac{0±\sqrt{\frac{196}{25}}}{2}$$

Take the square root of \frac{196}{25}.

$$a=\frac{0±\frac{14}{5}}{2}$$

Now solve the equation a=\frac{0±\frac{14}{5}}{2} when ± is plus.

$$a=\frac{7}{5}$$

Now solve the equation a=\frac{0±\frac{14}{5}}{2} when ± is minus.

$$a=-\frac{7}{5}$$

The equation is now solved.

$$a=\frac{7}{5} a=-\frac{7}{5}$$