Question

In simplest radical form, what are the solutions to the quadratic equation 0 = –3x2 – 4x + 5?

Answer 1

The solutions to the quadratic equation 0 = –3x² – 4x + 5 are x = -2/3 + √19/3 and x = -2/3 - √19/3, found using the quadratic formula.

Step-by-step explanation:

The solutions to the quadratic equation 0 = –3x² – 4x + 5 can be found using the quadratic formula, which is applicable for equations in the form ax² + bx + c = 0. The quadratic formula is given by x = (-b ± √(b² - 4ac)) / (2a). Applying this to our equation, we have:

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

First, we compute the discriminant (b² - 4ac):

(-4)² - 4(-3)(5) = 16 + 60 = 76

Next, we use the quadratic formula:

x = (4 ± √76) / (2 × -3)

x = (4 ± √(4² × 19)) / (-6)

x = (4 ± 2√19) / (-6)

x = (-2/3 ± √19/3)

So the solutions in simplest radical form are x = -2/3 + √19/3 and x = -2/3 - √19/3.

Answer 2

- 3 x² - 4 x + 5 = 0

`x_(1) = (4+ √(16+60) )/(-6) = (4+8.718)/(-6)`
x1 = -2.12

`x_(2) = (4-8.718)/(-6)`
x2= 0.786