Solve the equation t^2-2t+3=0
h = Xv = -B/2A = 2/2 = 1.
K = 1^2 - 2*1 + 3 = 2.
V(h, k) = V(1, 2).
For a parabola that opens upward:
k < 0: 2 real .solutions.
k = 0: 1 real solution.
k > 0: no real solutions.
there are no real number solutions
use the quadratic formula to solve
To solve the equation t^2 - 2t + 3 = 0, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a, b, and c represent the coefficients of the quadratic function. For the given equation, a = 1, b = -2, and c = 3.
Plugging these values into the quadratic formula, we get:
t = (-(-2) ± √((-2)^2 - 4(1)(3))) / (2(1))
Simplifying further:
t = (2 ± √(4 - 12)) / 2
t = (2 ± √(-8)) / 2
Now, we encounter a problem. The term √(-8) is not a real number because you cannot take the square root of a negative number within the set of real numbers. Thus, the equation t^2 - 2t + 3 = 0 has no real solutions.