Algebra quadratics will give brainlist
find the discriminant of the quadratic equation x2 + 6x + 14 = 0 and use it to determine the number and types of solutions.
b2 − 4ac
−20; two nonreal solutions
−20; one real solution
92; two real solutions
92; one real solution

A

Step-by-step explanation:

Given a quadratic equation in standard form : ax² + bx + c = 0

Then the discriminant Δ = b² - 4ac informs us about the nature of the roots.

• If b² - 4ac > 0 then 2 real, distinct roots

• If b² - 4ac = 0 then 2 real , equal roots

• If b² - 4ac < 0 then 2 non real roots

x² + 6x + 14 = 0 ← is in standard form

with a = 1, b = 6, c = 14, then

b² - 4ac = 6² - (4 × 1 × 14) = 36 - 56 = - 20

Since b² - 4ac < 0 then 2 non real roots → A

-20, two non real solutions.

Step-by-step explanation:

The general form of a quadratic is ax^2 + bx +c. In this case, a=1, b=6, and c=14. Plug in the values for the discriminant: b^2 - 4ac ---> (6)^2 - 4(1)(14) ---> 36 - 56 ---> -20.

Remember, if the discriminant is positive, then there are two real solutions. If the discriminant is 0, then there is one real solution. If the discriminant is negative there are two imaginary solutions.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation: