Answer
$x^2-4x+29=0$
Work Step by Step
The sum of the given roots, $
2+5i
\text{ and }
2-5i
,$ is
\begin{align*}
&
2+5i+2-5i
\\&=
(2+2)+(5i-5i)
\\&=
4+0i
\\&=
4
.\end{align*}
Since the sum is given by the ratio $-\dfrac{b}{a},$ then
\begin{align*}
-\dfrac{b}{a}&=4
\\\\
-\dfrac{b}{1}&=4
&\text{ (given $a=1$)}
\\\\
-b&=4
\\
b&=-4
.\end{align*}
The product of the given roots is
\begin{align*}\require{cancel}
&
(2+5i)(2-5i)
\\&=
2(2)+2(-5i)+5i(2)+5i(-5i)
&\text{ (use FOIL)}
\\&=
4-10i+10i-25i^2
\\&=
4-10i+10i-25(-1)
&\text{ (use $i^2=-1$)}
\\&=
4-10i+10i+25
\\&=
(4+25)+(-10i+10i)
\\&=
29+0i
\\&=
29
.\end{align*}
Since the product is given by the ratio $\dfrac{c}{a},$ then
\begin{align*}
\dfrac{c}{a}&=29
\\\\
\dfrac{c}{1}&=29
&\text{ (given $a=1$)}
\\\\
c&=29
.\end{align*}
Hence, the quadratic equation $ax^2+bx+c=0$ with the given roots is
\begin{align*}
x^2-4x+29=0
.\end{align*}
