Answer:for ax^2+bx+c=0 the discriminant is b^2-4ac
there are 3 basic cases of what happens for different discriminants
1. if the discriminant is less than 0, then there are no real zeroes
2. if the discriminant is 0, then it has 1 zero
3. if the discriminant is greater than 0, it has 2 zeroes
so given
0=3x^2-7x+4
a=3,b=-7,c=4
thus the discriminant is (-7)^2-4(3)(4)=49-48=1
the discriminant is 1. 1 is positive, thus the equation has 2 zeroes because the discriminant is greater than 0
the answer is the equation has two zeroes because the discriminant is greater than 0
Step-by-step explanation:
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Answer:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Step-by-step explanation:
So we have the expression:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
Recall the order of operations or PEMDAS:
P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.
E: Within the parentheses, if exponents are present, do them before all other operations.
M/D: Multiplication and division next, whichever comes first.
A/S: Addition and subtraction next, whichever comes first.
(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)
Thus, we should do the operations inside the parentheses first. Therefore:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
The parentheses is:

Square the 6 and the 4:

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

Therefore, the original equation is now:
![12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%5C%5C%3D12-%20%5B20-2%2848%29%5D)
Multiply with the brackets:
![=12-[20-96]](https://tex.z-dn.net/?f=%3D12-%5B20-96%5D)
Subtract with the brackets:
![=12-[-76]](https://tex.z-dn.net/?f=%3D12-%5B-76%5D)
Two negatives make a positive. Add:

Therefore:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)