-2x^2 + 5x - 3=0 ;;;;;;;;; (3.)3x^2 - 2x +6=0 ;;;;;; (5.) 6x^2 - 3x +4 =0?

Question

Find the discrimnant, and determine the number of real soultions. then solve .
please help me understand

Answer 1

Standard form of a quadratic equation:

ax^2 + bx + c = 0

Discriminant: b^2 - 4ac

If b^2 - 4ac > 0 there are two distinct real roots
If b^2 - 4ac = 0, there are two coincident real roots (one number)
If b^2 - 4ac < 0, there are two complex roots

In -2x^2 + 5x - 3 = 0, a = -2, b = 5, c = -3, so:

b^2 - 4ac = (5)^2 - 4(-2)(-3) = 25 - 24 = 1 > 0

Therefore, the equation has two real roots.

In 3x^2 - 2x +6 = 0, a = 3, b = -2, c = 6, so

b^2 - 4ac = (3)^2 - 4(3)(6) = 9 - 72 = -63 < 0

Therefore, there are two complex roots.

In 6x^2 - 3x + 4 = 0 , a = 6, b = -3, c = 4, so

b^2 - 4ac = (-3)^2 - 4(6)(4) = 9 - 96 = -87 < 0

Therefore there are two complex roots.

Answer 2

DISCRIMINANT IS B^2-4AC

IF THE DISCRIMINANT IS POSITIVE YOU HAVE TWO REAL SOLUTIONS
IF THE DISCRIMINANT IS ZERO YOU HAVE ONE REAL SOLUTION
IF THE DISCRIMINATN IS NEGATIVE THEN YOU HAVE NO REAL SOLUTIONS

IN THE FIRST EQUATION A=-2 B=5 C=-3
YOUR DISCRIMINANT WOULD BE 5^2 - 4X(-2)X(-3)
which is 25-24 = 1 first equation has a discriminant of 1 which is postive which means two real solutions

2nd equation
a=3 b=-2 c=6
discriminant is -2^2 - 4X(3)X(6)
which is 4 - 72 which is -68 which is negative which means you have two complex solutions - no real solutions

3rd equation
a=6 b= -3 c = 4
discriminant is (-3)^2 - 4X(6)X(4)
which is 9 - 96 which is -85 which means you have two complex solutions - no real solutions

Answer 3

Question 1
2x² - 5x + 3 = 0
x = [ 5 ± √(25 - 24) ] /4
x = [ 5 ± 1] / 4
x = 3/2 , x = 1 (two real solutions)

Question 2
x = [ 2 ± √(4 - 72) ] / 6
x = [ 2 ± √(- 68) ] / 6
x = [ 2 ± i √(68) ] / 6
x = [ 2 ± 2i √17 ] / 4
x = (1/2) [ 1 ± i √17 ] (imaginary solutions)

Question 3
x = [ 3 ± √(9 - 96) ] / 12
x = [ 3 ± √ 90 ] /12
x = [ 3 ± 3√10 ] / 12
x = (1/4).(1 ± √10 ) (two real solutions)