using the quadratic formula to solve $x^{2}+20 = 2x$, what are the values of $x$?\n$1pmsqrt{21}i$\n$-1pmsqrt{…

Explanation:

Step1: Rewrite in standard form

First, rewrite the equation $x^{2}+20 = 2x$ as $x^{2}-2x + 20=0$. For a quadratic equation $ax^{2}+bx + c = 0$, here $a = 1$, $b=-2$, $c = 20$.

Step2: Apply quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 1$, $b=-2$, $c = 20$ into it: [ \begin{align*} x&=\frac{-(-2)\pm\sqrt{(-2)^{2}-4\times1\times20}}{2\times1}\ &=\frac{2\pm\sqrt{4 - 80}}{2}\ &=\frac{2\pm\sqrt{- 76}}{2}\ &=\frac{2\pm2\sqrt{19}i}{2} \end{align*} ]

Step3: Simplify the result

Simplify $\frac{2\pm2\sqrt{19}i}{2}$ to get $x = 1\pm\sqrt{19}i$.

Answer:

$1\pm\sqrt{19}i$