Question

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

Answer 1

Answer:

x = - $\frac{2}{3}$ ± $\frac{\sqrt{19} }{3}$

Step-by-step 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 ± $\sqrt{(-4)^2-(4(-3)(5)}$) / - 6

  = ( 4 ± $\sqrt{16+60}$) / - 6

  = ( 4 ± $\sqrt{76}$) / - 6

  = (4 ± 2$\sqrt{19}$) / - 6

  = - $\frac{2}{3}$ ± $\frac{\sqrt{19} }{3}$