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solve for the solutions of (x^{2}-4x + 13=0). use the keypad to enter y…

Question

solve for the solutions of (x^{2}-4x + 13=0). use the keypad to enter your answers in the boxes. additional symbols can be found using the drop - down arrows at the top of the keypad. the solutions are (x=square) and (x=square).

Explanation:

Step1: Identify quadratic coefficients

For $ax^2+bx+c=0$, here $a=1$, $b=-4$, $c=13$

Step2: Calculate discriminant

$\Delta = b^2-4ac = (-4)^2 - 4(1)(13) = 16 - 52 = -36$

Step3: Apply quadratic formula

$x = \frac{-b\pm\sqrt{\Delta}}{2a} = \frac{4\pm\sqrt{-36}}{2}$

Step4: Simplify complex root

$\sqrt{-36}=6i$, so $x=\frac{4\pm6i}{2}=2\pm3i$

Answer:

$x=2+3i$ and $x=2-3i$