Solve 16b^2-25=0

$$16 b ^ { 2 } - 25 = 0$$

Solve for b

$$b = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$b = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Steps Using Difference of Squares

Consider 16b^{2}-25. Rewrite 16b^{2}-25 as \left(4b\right)^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4b-5\right)\left(4b+5\right)=0$$

To find equation solutions, solve 4b-5=0 and 4b+5=0.

$$b=\frac{5}{4} b=-\frac{5}{4}$$

Steps by Finding Square Root

Add 25 to both sides. Anything plus zero gives itself.

$$16b^{2}=25$$

Divide both sides by 16.

$$b^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$b=\frac{5}{4} b=-\frac{5}{4}$$

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.

$$16b^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times 16\left(-25\right)}}{2\times 16}$$

Square 0.

$$b=\frac{0±\sqrt{-4\times 16\left(-25\right)}}{2\times 16}$$

Multiply -4 times 16.

$$b=\frac{0±\sqrt{-64\left(-25\right)}}{2\times 16}$$

Multiply -64 times -25.

$$b=\frac{0±\sqrt{1600}}{2\times 16}$$

Take the square root of 1600.

$$b=\frac{0±40}{2\times 16}$$

Multiply 2 times 16.

$$b=\frac{0±40}{32}$$

Now solve the equation b=\frac{0±40}{32} when ± is plus. Reduce the fraction \frac{40}{32} to lowest terms by extracting and canceling out 8.

$$b=\frac{5}{4}$$

Now solve the equation b=\frac{0±40}{32} when ± is minus. Reduce the fraction \frac{-40}{32} to lowest terms by extracting and canceling out 8.

$$b=-\frac{5}{4}$$

The equation is now solved.

$$b=\frac{5}{4} b=-\frac{5}{4}$$