3v^2+4v-1=0

Question

Find all real or imaginary solutions to the equation

Answer 1

3v² + 4v - 1 = 0

 

x = [-b ± √(b² - 4ac)] / 2a, wheree a = 3, b = 4, c = -1

x = [-4 ± √(-4² - 4(3)(-1))] / 2(3)

x = [-4 ± √(16 + 12)] / 6

x = [-4 ± √28] / 6

x = -2/3 ± √28/6