Question

Solve by using the Quadratic Formula:$4z^2+2z-6=0$

Short answer

The roots of the above equation is$z=1$and$z=-\frac{3}{2}$

Step-by-step solution

Given information

We are given a quadratic equation$4z^2+2z-6=0$

Step 2: Identify the values of a,b,c

On comparing with standard form we get the values of$a,b,c$as$a=4,b=2,c=-6$

Write the quadratic equation formula and substitute the values of a,b,c and then simplify and solve for x

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

Substituting the values we get

$$\begin{gathered} z=\frac{-2\pm \sqrt{4-4\left(4\right)(-6)}}{2\left(4\right)} \\ z=\frac{-2\pm \sqrt{4+96}}{8} \\ z=\frac{-2\pm \sqrt{100}}{8} \\ z=\frac{-2\pm 10}{8} \end{gathered}$$

On further simplifying we get

$$\begin{gathered} z=\frac{-2+10}{8} \\ z=\frac{8}{8} \\ z=1 \end{gathered}$$and$$\begin{gathered} z=\frac{-2-10}{8} \\ z=\frac{-12}{8} \\ z=-\frac{3}{2} \end{gathered}$$

Step 4:  check the solutions

We substitute $x=1$in the quadratic equation

$$\begin{gathered} 4z^2+2z-6 \\ 4{\left(1\right)}^2+2\left(1\right)-6 \\ 4+2-6 \\ =0 \end{gathered}$$

Now we substitute $z=-\frac{3}{2}$in the above equation

$$\begin{gathered} 4z^2+2z-6 \\ 4{(-\frac{3}{2})}^2+2(-\frac{3}{2})-6 \\ 9-3-6 \\ =0 \end{gathered}$$

Conclusion

The solution of the above equation is$z=1$and$z=-\frac{3}{2}$