t = \frac{-b}{2a} = \frac{-30}{2(-5)} = \frac{-30}{-10} = 3

["Finding the Vertex Using the Formula: A Step-by-Step Guide", "In algebra, identifying the vertex of a quadratic equation is essential for understanding its graph and behavior. One of the most direct ways to find the x-coordinate of the vertex is by using the formula ( t = \frac{-b}{2a} ). This formula is derived from completing the square and provides a quick shortcut when working with equations in the standard form:\n[ f(t) = at^2 + bt + c ]", "Let’s break down an essential calculation:\nIf we have the quadratic function\n[\nf(t) = -5t^2 - 30t + 10\n]\nthe vertex occurs at\n[\nt = \frac{-b}{2a}\n]", "Substituting ( a = -5 ) and ( b = -30 ):\n[\nt = \frac{-(-30)}{2(-5)} = \frac{30}{-10} = -3\n]", "Wait—here, we see a common pitfall: the negatives cancel differently depending on signs. Recheck carefully:\n[\n\frac{-b}{2a} = \frac{-(-30)}{2(-5)} = \frac{30}{-10} = -3\n]", "But let’s verify again: if ( b = -30 ), then ( -b = 30 ).\nSince ( a = -5 ) is negative, ( 2a = -10 ).\nSo:\n[\nt = \frac{30}{2(-5)} = \frac{30}{-10} = -3\n]", "Wait—this yields ( t = -3 ), not 3. But where does ( \frac{-30}{2(-5)} = \frac{-30}{-10} = 3 ) come from?", "Ah—here’s the key: in the original expression,\n[\n\frac{-b}{2a}, \quad b = -30 \Rightarrow -b = 30, \quad a = -5\n]\n[\nt = \frac{30}{2(-5)} = \frac{30}{-10} = -3\n]", "So the claim that ( \frac{-30}{2(-5)} = 3 ) is incorrect—unless the original equation had ( b = 30 ), not ( b = -30 ). Let's carefully reinterpret the example:", "Suppose the quadratic equation is:\n[\nf(t) = -5t^2 + 30t - 10\n]\nThen ( a = -5 ), ( b = 30 ), and\n[\nt = \frac{-b}{2a} = \frac{-30}{2(-5)} = \frac{-30}{-10} = 3\n]", "Yes—now the result is 3. This matches your result perfectly.", "---", "### Why is This Formula Important?", "The x-coordinate of the vertex, ( t = \frac{-b}{2a} ), gives the axis of symmetry for the parabola defined by ( f(t) = at^2 + bt + c ). Using this value allows us to determine the maximum or minimum point (depending on the sign of ( a )), vertex form, and key graph features.", "For example, in ( f(t) = -5t^2 + 30t - 10 ), plugging ( t = 3 ) back into the equation finds the y-coordinate:\n[\nf(3) = -5(3)^2 + 30(3) - 10 = -45 + 90 - 10 = 35\n]", "Thus, the vertex is at ( (3, 35) ), confirming the symmetry and peak of the parabola opens downward (since ( a = -5 < 0 )).", "---", "### Summary: How to Calculate the Vertex Using ( t = \frac{-b}{2a} )", "1. Identify coefficients ( a ), ( b ), and ( c ) from the quadratic equation.\n2. Substitute into the formula:\n [\n t = \frac{-b}{2a}\n ]\n3. Simplify carefully with proper sign handling.\n4. Use ( t ) as the x-coordinate of the vertex; substitute into the original equation to find the y-coordinate.", "This method saves time compared to completing the square and is foundational for graphing and analyzing quadratic functions.", "---", "Conclusion:\nUnderstanding and applying ( t = \frac{-b}{2a} ) is crucial in algebra. While sign errors can change the result, carefully evaluating each step ensures accuracy. When ( b = 30 ) and ( a = -5 ), the x-coordinate of the vertex is indeed 3—demonstrating how formula application and sign awareness combine to unlock deeper insights into quadratic behavior.", "---", "Keywords: vertex formula, ( t = \frac{-b}{2a} ), quadratic vertex, algebra, solve quadratic, mathematical formula, vertex of a parabola, graphing quadratics, completing the square alternatively, algebraic steps."]