An algebra problem by Wildan Bagus Wicaksono

$\begin{aligned} \sqrt{2017x^2+4034x+17} + \sqrt{2017x^2+4034x-3} & = 10 & \small \color{#3D99F6} \text{Let }u = 2017x^2 + 4034x + 7 \\ \sqrt{u+10} + \sqrt{u-10} & = 10 & \small \color{#3D99F6} \text{Squaring both sides.} \\ u+10 + 2\sqrt{u^2-100} + u - 10 & = 100 \\ 2\sqrt{u^2-100} & = 100 - 2u \\ \sqrt{u^2-100} & = 50 - u & \small \color{#3D99F6} \text{Squaring both sides.} \\ u^2 - 100 & = u^2 - 100u + 2500 \\ u & = 26 \\ \implies 2017x^2 + 4034x + 7 & = 26 \\ 2017x^2 + 4034x & = 19 \\ \implies x^2 + 2x & = \boxed{\dfrac {19}{2017}} \end{aligned}$

$\large \sqrt { 2017{ x }^{ 2 }+4034x+17 } =p\\ \large \sqrt { 2017{ x }^{ 2 }+4034x-3 } =q$

So, $p+q=10$

$\large 10(p-q)=(p+q)(p-q)\\ \large 10(p-q)={ p }^{ 2 }-{ q }^{ 2 }\\ \large 10(p-q)={ \left( \sqrt { 2017{ x }^{ 2 }+4034x+17 } \right) }^{ 2 }-{ \left( \sqrt { 2017{ x }^{ 2 }+4034x-3 } \right) }^{ 2 }\\ \large 10(p-q)=2017{ x }^{ 2 }+4034x+17-(2017{ x }^{ 2 }+4034x-3)\\ \large 10(p-q)=2017{ x }^{ 2 }+4034x+17-{ 2017x }^{ 2 }-4034x+3\\ \large10(p-q)=20\\ p-q=2$

$\large p+q=10\\ \large p-q=2\\ -------+\\ \large 2p=12\\ \large p=6$

So we get,

$\large p=6\\ \large \sqrt { 2017{ x }^{ 2 }+4034x+17 } =6\\ \large 2017{ x }^{ 2 }+4034x+17=36\\ \large 2017({ x }^{ 2 }+2x)=36-17\\ \large 2017({ x }^{ 2 }+2x)=19\\ \large { x }^{ 2 }+2x=\frac { 19 }{ 2017 }$

$\sqrt{2017x^2+4034x+17}+\sqrt {2017x^2+4034x-3}=10$

$\sqrt{2017(x^2+2x)+17}+\sqrt{2017 (x^2+2x)-3}=10$

$\sqrt{2017(x^2+2x)+17}=A,\sqrt{2017 (x^2+2x)-3}=B$

$\text{Keeping in mind that},A+B=10$

$A^2-B^2=10(A-B)$

$\therefore (\sqrt{2017(x^2+2x)+17})^2-(\sqrt {2017 (x^2+2x)-3})^2=10(\sqrt{2017(x^2+2x)+17}-\sqrt {2017 (x^2+2x)-3})$

$\therefore 2017 (x^2+2x)+17-2017 (x^2+2x)+3=10 (\sqrt {2017 (x^2+2x)}-\sqrt {2017 (x^2+2x)-3})$

$\therefore 20=10 (A-B)$

$\therefore A-B=2$

$\text{Since A+B=10 and A-B=2, A=6 and B=4}$

$\therefore \sqrt {2017 (x^2+2x)+17}=A=6$

$\therefore 2017 (x^2+2x)+17=36$

$\therefore 2017 (x^2+2x)=19$

$\therefore x^2+2x=\frac {19}{2017}$