Question

In simplest radical form, what are the solutions to the quadratic equation 0 = –3x^2 – 4x + 5?

Answer 1

x = -
`(2)/(3)` ±
`(√(19) )/(3)`

Explanation:

given a quadratic equation in the form ax² + bx + c = 0

We can solve using the quadratic formula

with a = - 3, b= - 4 and c = 5

x = ( 4 ±
`√((-4)^2-(4(-3)(5))`) / - 6

= ( 4 ±
`√(16+60)`) / - 6

= ( 4 ±
`√(76)`) / - 6

= (4 ± 2
`√(19)`) / - 6

= -
`(2)/(3)` ±
`(√(19) )/(3)`