Answer
$\{9,11\}$ or $\{-11,-9\}$
Work Step by Step
Two consecutive odd integers, which can be represented by $x$ and $(x+2)$, are squared then added together to equal $202$
$(x)^2 + (x+2)^2=202$
$x^2 + x^2+4x+4=202$
$2x^2+4x+4=202$
$2x^2+4x-198=0$
Now, apply the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{-4\pm\sqrt{4^2-4(2)(-198)}}{2(2)}$
$x=\dfrac{-4\pm\sqrt{16-8(-198)}}{4}$
$x=\dfrac{-4\pm\sqrt{16-8(-198)}}{4}$
$x=\dfrac{-4\pm\sqrt{16+1584}}{4}$
$x=\dfrac{-4\pm\sqrt{1600}}{4}$
$x=\dfrac{-4\pm40}{4}$
$x=\dfrac{-4+40}{4}$ or $x=\dfrac{-4-40}{4}$
$x=\dfrac{36}{4}$ or $x=\dfrac{-44}{4}$
$x=9$ or $x=11$
Now we check the answer:
$9^2 +11^2 =202$ and $(-9)^2 +(-11)^2 =202$
$81 +121 =202$
So the solutions are $\{9,11\}$ or $\{-11,-9\}$
