Question

How do you solve `3z ^ { 2} + 4z - 7= 0`?

Answer 1

`z=-7/3 or z = 1`

Explanation:

Using the quadratic equation

`z_(1,2) = (-b +- sqrt(b^2 –4ac))/(2a)`

you get

`z= (-4 +- sqrt( 4^2 -4xx3xx (-7)))/(2xx3)`

`z= (-4 +- sqrt(16+84))/6`

`z= (-4 +- sqrt(100))/6`

`z= (-4 +- 10)/6`

`z= -14/6 or z = 6/6`

`z=-14/6 or z = 1`

`z= -7/3 or z = 1`

Answer 2

1, and `-7/3`

Explanation:

`3z^2 + 4z - 7 = 0`
Use shortcut when a + b + c = 0
`z = 1`, and `z = c/a = - 7/3`

Reminder of shortcut --> `y = ax^2 + bx + c = 0`

1. When a + b + c = 0
   `x1 = 1`, and `x2 = c/a`
   Example: `y = 3x^2 + 4x - 7 = 0`
   `x1 = 1`, and `x2 = - 7/3`

2. When a - b + c = 0.
   x1 = - 1, and `x2 = - c/a`
   Example. `y = 5x^2 + 3x - 2x = 0`
   `x1 = - 1`, and `x2 = -c/a = 2/5`