Question

In the following exercises, solve by using the Quadratic Formula.
$\frac{3}{4}{b}^2+\frac{1}{2}b=\frac{3}{8}$

Short answer

The solutions of the given equation are $b=-\frac{1}{3}\pm \frac{\sqrt{22}}{6}$.

Step-by-step solution

Step 1. Write the quadratic equation in standard form.

The given equation is:

$\frac{3}{4}{b}^2+\frac{1}{2}b=\frac{3}{8}$

The given equation can be written as:

$\begin{array}{rcl}\frac{3}{4}{b}^2+\frac{1}{2}b-\frac{3}{8}& =& 0\\ \frac{6b^2+4b-3}{8}& =& 0\\ 6b^2+4b-3& =& 0\end{array}$

Step 2. Substitute the values in Quadratic Formula.

Quadratic Formula for the equation $a{x}^2+bx+c=0$is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In the given equation, $a=6,b=4,c=-3$. Substitute these values in Quadratic Formula.

$b=\frac{-4\pm \sqrt{(4{)}^2-4(6\left)\right(-3)}}{2\left(6\right)}$

Step 3. Simplify the value.

On simplification, we get

$\begin{array}{rcl}b& =& \frac{-4\pm \sqrt{16+72}}{12}\\ b& =& \frac{-4\pm \sqrt{88}}{12}\\ b& =& \frac{-4}{12}\pm \frac{2\sqrt{22}}{12}\\ b& =& -\frac{1}{3}\pm \frac{\sqrt{22}}{6}\end{array}$

Step 4. Conclusion.

The solutions of the given equation are $b=-\frac{1}{3}\pm \frac{\sqrt{22}}{6}$.