t^2+80t+300=0

asked by guest
on Nov 14, 2024 at 3:03 am



You asked:

Solve the equation \({t}^{2} + 80 t + 300 = 0\) for the variable \(t\).

MathBot Answer:

\[\begin{aligned}t &= -40 + 10 \sqrt{13} \approx -3.9444872\\t &= -40 - 10 \sqrt{13} \approx -76.055513\end{aligned}\]

Solve by quadratic formula:

Given a quadratic equation \(a x^{2} + b x + c = 0\), where \(a\), \(b\), \(c\) are constants and \(a \ne 0\), the solutions are given by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\] In this case \(a=1\), \(b=80\), and \(c=300\).

The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:\[b^{2}-4ac = 80^{2} - 4 \cdot 300=5200 > 0\] The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: \[t = \frac{\left(-1\right) 80 + \sqrt{5200}}{2 \cdot 1} = -40 + 10 \sqrt{13} \approx -3.9444872\] \[t = \frac{\left(-1\right) 80 - \sqrt{5200}}{2 \cdot 1} = -40 - 10 \sqrt{13} \approx -76.055513\]

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