x^2 +(2x+5)^2 +(7x+8)^4 =0

Pascals triangle 1 1 1 1 2 1 1 3 3 1                      1 4 6 4 1

$$\\(7x+8)^4\\
=(7x)^4+4*(7x)^3*8+6*(7x)^2*8^2+4*(7x)*8^3+8^4\\
=2401x^4+4*343x^3*8+6*49x^2*64+4*7x*512+4096\\
=2401x^4+10976x^3+18816x^2+14336x+4096\\

\begin{array}{rll}
x^2 +(2x+5)^2 +(7x+8)^4 &=&0\\
x^2 +4x^2+20x+25 +2401x^4+10976x^3+18816x^2+14336x+4096 &=&0\\
2401x^4+10976x^3+18821x^2+14356x+4121 &=&0\\
\end{array}$$

Now I have lost interest.

Here is your answer;

http://www.wolframalpha.com/input/?i=x%5E2+%2B%282x%2B5%29%5E2+%2B%287x%2B8%29%5E4+%3D0

There are no real solutions :)

Here's a "close-up" picture....Melody is correct.....no real solutions and (surprisingly) only one turning point.......https://www.desmos.com/calculator/ks0cm9nazp

Thanks Chris