Please use " ^ " to indicate exponentiation: x^2 + 3x - 4 = 0
Here, a = 1, b = 3 and c = -4.
The formula for the discriminant is b^2 - 4(a)(c).
Substituting the given values of a, b and c, we get:
(3)^2 - 4(1)(-4)
Evaluating this, we get 9 + 16 = 25.
The discriminant is 25.
-3 plus or minus √25
Taking this further, x = ------------------------------------
2
-3 plus or minus 5
or: x = -------------------------------- => {-4, 1} (solutions)
2
In an arithmetic sequence, the difference between consecutive terms is constant. In formulas, there exists a number
such that

In an geometric sequence, the ratio between consecutive terms is constant. In formulas, there exists a number
such that

So, there exists infinite sequences that are not arithmetic nor geometric. Simply choose a sequence where neither the difference nor the ratio between consecutive terms is constant.
For example, any sequence starting with

Won't be arithmetic nor geometric. It's not arithmetic (no matter how you continue it, indefinitely), because the difference between the first two numbers is 14, and between the second and the third is -18, and thus it's not constant. It's not geometric either, because the ratio between the first two numbers is 15, and between the second and the third is -1/5, and thus it's not constant.
The answer is D
Let me know if it’s right!
There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10
[2] b-c=15
[3] a-2b+c=-5
Equations Simplified or Rearranged :<span><span> [1] 2a + c = -10
</span><span> [2] - c + b = 15
</span><span> [3] a + c - 2b = -5
</span></span>Solve by Substitution :
// Solve equation [3] for the variable c
<span> [3] c = -a + 2b - 5
</span>
// Plug this in for variable c in equation [1]
<span><span> [1] 2a + (-a +2?-5) = -10
</span><span> [1] a = -5
</span></span>
// Plug this in for variable c in equation [2]
<span><span> [2] - (-? +2b-5) + b = 15
</span><span> [2] - b = 10
</span></span>
// Solve equation [2] for the variable ?
<span> [2] ? = b + 10
</span>
// Plug this in for variable ? in equation [1]
<span><span> [1] (? +10) = -5
</span><span> [1] 0 = -15 => NO solution
</span></span><span>No solution</span>