3x^2-4x-5=0?

Question

Use the quadratic formula to solve. Leave irrational roots in the simplest radical form.

Answer 1

Quadratic Formula

x = - b ± √b² - 4ac / 2a

3x² - 4x - 5

Let

a = 3,

b = - 4

c = - 5

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

x = 4 ± √16 - (- 60) / 6

x = 4 ± √16 + 60 / 6

x = 4 ± √76 / 6

x = 4 ± 8.717797887 / 6

- - - - - - - - - -

Solving for +

x = 4 + 8.717797887 / 6

x = 12.71779789 / 6

x = 2.119632981

- - - - - -

solving for -

x = 4 -8.717797887 / 6

x = - 4.717797887 / 6

x = - 0.786299648

- - - - - - - -s-

Answer 2

Ans.

For an eqn ax^2 + bx + c = 0.

Roots are given by
x = [-b +/- SQRT (b^2 - 4ac)]/2a

Here, a = 3, b = -4, c = -5

Substituting we get,
x = [4 +/- SQRT (16 + 60)]/6
x = [2 + SQRT(19)]/3, [2 - SQRT(19)]/3

Answer 3

(-2 +or- 2 times sq. root of 19) divided by 3

Answer 4

x = [4 ±√(16 + 60)] / 6
x = [4 ±√(76)] / 6
x = [4 ± 2√19] / 6
x = (1/3).[2 ± √19]

Answer 5

x = -(-4) +/- sqr(16 - 4 x 3 x -5) divided by 6

x = 4 +/- sqr(76) divided by 6
x = 4 +/- 2 sqr(19) divided by 6

x = {2 +/- sqr(19)} div by 3