Question

solve the equation using the quadratic formula.
$x^{2}-4x + 13 = 0$
the solution set is
(simplify your answer. type an exact answer, using radicals and $i$ as needed. use integers numbers in the expression. use a comma to separate answers as needed.)

Explanation:

Step1: Identify coefficients

For the quadratic equation $x^{2}-4x + 13=0$, we have $a = 1$, $b=-4$, $c = 13$.

Step2: Write quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step3: Substitute coefficients

Substitute $a = 1$, $b=-4$, $c = 13$ into the formula:
$x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4\times1\times13}}{2\times1}=\frac{4\pm\sqrt{16 - 52}}{2}=\frac{4\pm\sqrt{- 36}}{2}$.

Step4: Simplify square - root

Since $\sqrt{-36}=\sqrt{36\times(-1)} = 6i$, then $x=\frac{4\pm6i}{2}$.

Step5: Simplify the fraction

$x=\frac{4}{2}\pm\frac{6i}{2}=2\pm3i$.

Answer:

$2 + 3i,2 - 3i$