-12x^(2)-tx+t=0 In the given equation, t is a nonzero constant. If the equation has only one real solution, s, what is the value of s ?

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-12x^(2)-tx+t=0 In the given equation,  t is a nonzero constant. If the equation has only one real solution,  s, what is the value of  s ?

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Understand Conditions for Real Solution: To find the value of the real solution s, we need to understand the conditions for a quadratic equation to have only one real solution. A quadratic equation ax^2 + bx + c = 0 has only one real solution when its discriminant is zero. The discriminant is given by the formula b^2 - 4ac.Calculate Discriminant: In the given equation -12x^2 - tx + t = 0, we can identify a = -12, b = -t, and c = t. Let's calculate the discriminant using these values. Discriminant (D) = b^2 - 4ac = (-t)^2 - 4(-12)(t) = t^2 + 48t.Set Discriminant Equal to Zero: For the equation to have only one real solution, the discriminant must be zero. Therefore, we set the discriminant equal to zero and solve for t. t^2 + 48t = 0Factorize and Solve for t: We can factor out t from the equation: t(t + 48) = 0Find Value of t: Since t is a nonzero constant, we cannot have t = 0. Therefore, the only solution for t is when t + 48 = 0, which gives us t = -48.Calculate Vertex for Real Solution: Now that we have the value of t, we can find the value of the real solution s. Since the equation has only one real solution, this solution is also the vertex of the parabola represented by the quadratic equation. The x-coordinate of the vertex is given by -b/(2a).Substitute a = -12 and b = -t (with t = -48) into the vertex formula to find s: s = -(-t)/(2a) = -(-(-48))/(2(-12)) = 48/(-24) = -2.

$-12 x^{2}-t x+t=0$ $\newline$ In the given equation, $t$ is a nonzero constant. If the equation has only one real solution, $s$ , what is the value of $s$ ?

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−12x2−tx+t=0 -12 x^{2}-t x+t=0 −12x2−tx+t=0\newlineIn the given equation, t t t is a nonzero constant. If the equation has only one real solution, s s s, what is the value of s s s ?

$-12 x^{2}-t x+t=0$ $\newline$ In the given equation, $t$ is a nonzero constant. If the equation has only one real solution, $s$ , what is the value of $s$ ?