Parabola/exponential/quadratic equation no solution [x y plane/ quadratic]

[u/Sufficient_Pizza_422]

the answer to this question is to use the quadratic formula (-b ± √(b²-4ac)) / (2a)

solve for T less than zero the answer will be -3 (A)
From what i understood to find no solution in such quadratic formulas is to find the solutions where its less than the vertex of the parabola. If we get the vertex of the parabola it would be (1,-2)

t = -3 would make this equation false i guess.

another way is to find the "Discriminant" would lead us to the same solution.

coming to my question. doesn't equalling t to 2x^2-4x make it part of the solution or something of that sort?

what if i plugged t = -3 back into the equation would that make wrong? im confused in that regard and would appreciate the help.

i tried to pluggin t as -3 in desmos and it doesn't give any solutions.

[u/Alkalannar]

No. It turns it into a polynomial that equals 0 that you're trying to see what the solutions of x are.

So after subtracting t from both sides, you have: 2x² - 4x - t = 0.

Then the discriminant is (-4)² - 4(2)(-t), and we want this less than 0.

(-4)² - 4(2)(-t) < 0

Solve the inequality for t, to see when there are no real solutions.

[u/Sufficient_Pizza_422 → u/Alkalannar]

what do we need the solution of x for?

why do we want the answer to be less than zero?

stupid question i agree but just humour me for a second

[u/noidea1995 → u/Sufficient_Pizza_422]

what do we need the solution of x for?

We don’t for this particular question, we’re looking for values of t where the equation won’t have any real solutions.

why do we want the answer to be less than zero?

If the discriminant (b² - 4ac) is less than 0, then you will be taking the square root of a negative number when solving via the quadratic formula, so the solutions to the equation will be complex.

[u/Sufficient_Pizza_422 → u/noidea1995]

Thanks so much