Question

solve the equation: $t^2 + 4t + 27 = 0$. fully simplify all answers, including non - real solutions.
t =
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Explanation:

Step1: Identify coefficients for quadratic formula

For the quadratic equation \(at^{2}+bt + c = 0\), here \(a = 1\), \(b = 4\), \(c = 27\). The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate the discriminant

First, calculate the discriminant \(D=b^{2}-4ac\). Substitute the values: \(D = 4^{2}-4\times1\times27=16 - 108=- 92\).

Step3: Simplify the square root of discriminant

Since \(\sqrt{-92}=\sqrt{4\times(-23)} = 2\sqrt{-23}=2i\sqrt{23}\) (using \(i=\sqrt{-1}\)).

Step4: Apply quadratic formula

Now substitute into the quadratic formula: \(t=\frac{-4\pm2i\sqrt{23}}{2\times1}=\frac{-4\pm2i\sqrt{23}}{2}=-2\pm i\sqrt{23}\).

Answer:

\(t=-2 + i\sqrt{23}\) or \(t=-2 - i\sqrt{23}\)