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

Question

How do I find the value of the discriminant of the equation, describe the roots completely, but not solve.

Answer 1

The discriminant is the value of B^2 - 4AC. This is the value under the square root of the quadratic formula.
a=3, b=4, c=-5
4^2 - (4)(3)(-5) = 76

Since the discriminant is positive, the roots are real.
sqrt(76)
sqrt(76)/2a

The roots must be symmetric about some value. That value is -B/(2A)
-4 / (2*3) = -2/3

Your roots will be
-2/3 + sqrt(76)/2a and -2/3 - sqrt(76)/2a

1/3 and -5/3

Answer 2

x = [ - 4 ± √(16 + 60) ] / 6
x = [ - 4 ± √(76) ] / 6
x = [ - 4 ± 2√(19) ] / 6
Will not solve, as requested.

Answer 3

Use the quadratic formula

ax^2 + bx + c = 0
when
x = [-b +- Sq Rt (b^2 -4ac)]/2a

3x^2 + 4x - 5 = 0
x = [-4 +- Sq Rt (16 + 60)]/6
x = (-4 +- 8.72)/6
x = 0.786
x = - 2.12

Check:

3(0.786)^2 +4(0.786) - 5 = 0
1. 853 + 3.144 - 5 = 0

3(-2.12)^2 + 4(-2.12) - 5 = 0
13.483 - 8.48 - 5 = 0

Answer 4

the descriminant is the sqrt (b^2-4ac)

so in this one it is the sqrt(4^2-4*3*(-5)) = sqrt(76)

So it has positive real roots

Answer 5

3x^2+4x-5=0
a=3, b=4, c=-5
d=16+60=76
x=(-4+sqrt(76))/6 or (-4-sqrt(76))/6

Answer 6

Discriminent is sq(b^2-4ac)
i,e.,sq(4^2-4.3.(-5))
i.e.,sq(16+60)
i.e.,Sq(76)
So the roots are real.