Question

How do you solve `-6z^2 + 7z + 3 = 0` using the quadratic formula?

Answer 1

`z = -0.3333 and z = 1.5`

Explanation:

Standard Quadratic Equation:

`x^2 + bx + c = 0`

Quadratic formula:

`(-b+-sqrt(b^2 - 4ac)) / (2a)`

b = constant with variable to the power 1
a = constant with variable to the power squared
c = constant.

`-6z^2 + 7z + 3 = 0`
In this case:

a = -6
b = 7
c= 3

put all these values

`(-7 +- sqrt((7)^2 - 4(-6)(3))) / (2(-6))`
`(-7 +- sqrt(49 + 72)) / -12`
`(-7 +- sqrt(121)) / -12`
`(-7 +- 11) / -12`
`(-7 + 11) / -12 , (-7 - 11) / -12`
`4 / -12, -18 / -12`
`z = -0.3333 and z = 1.5`