Finding the sum of real solutions to an equation how to find the sum of real solutions 
if,
$(x+1)(x+\frac14)(x+\frac12)(x+\frac34)=\frac{45}{32}$
I have tried multiplying both sides with 32 and got
$\frac{1}{32}(x+1)(2x+1)(4x+1)(4x+3)=\frac{45}{32}$
then i multiplied and got
$\frac{1}{32}(32x^4+80x^3+70x^2+25x+3)=\frac{45}{32}$
which is equal to
$32x^4+80x^3+70x^2+25x+3=45$
and
$32x^4+80x^3+70x^2+25x-42=0$
but i couldn't come up with something else...
Any help is appreciated!
 A: $$(x+1)(x+\frac{1}{4})(x+\frac{2}{4})(x+\frac{3}{4})=\frac{(4x+4)(4x+3)(4x+2)(4x+1)}{64}=\frac{45}{32}\\(4x+4)(4x+3)(4x+2)(4x+1)=90\\t=4x+\frac{5}{2}\\(t+\frac{3}{2})(t+\frac{1}{2})(t-\frac{1}{2})(t-\frac{3}{2})=90\\(t^2-\frac{9}{4})(t^2-\frac{1}{4})=90\\t^4-\frac{10}{4}t^2+\frac{9}{4}=90\\4t^4-10t^2-351=0$$
Writing $r=t^2$,Solving this quadratic you get that $r_1=\frac{5}{4}-\sqrt{91}$ and $r_2=\frac{5}{4}+\sqrt{91}$ since $r_1<0$ the solutions $t_1,t_2$ are complex hence we're only looking at $t_3,t_4$,since $t_3=-t_4$ we have that $$t_3+t_4=0\\4x_3+\frac{5}{2}+4x_4+\frac{5}{2}=0\\4(x_3+x_4)=-5\\x_3+x_4=-\frac{5}{4}$$
A: This is a fine start. Now you should know something about the sum of roots of a polynomial, right ? In case you don't: write your polynomial
$ x^4 + 5/2 x^2 + 35/16 x^2 + 25/32 x - 21/16 = 0$ as $(x-a)(x-b)(x-c)(x-d)$. Expand the right-hand side. What can you identify with $a+b+c+d$ ?
A: If all the solutions are real, it's simple. If 
$$P(x) = (x-x_1)(x-x_2)\cdots(x-x_n) = \sum_{k=0}^n a_k x^k$$
Then 
$$a_{n-1} = -\sum_{k=0}^n x_k$$
It become more complicated if only some solutions are real
A: $(x+1)(x+\frac14)(x+\frac12)(x+\frac34)=\frac{45}{32}$
$(x+1)(4x+1)(2x+1)(4x+3)=45$
$(4x^2+5x+1)(8x^2+10x+3)=45$
if we multiply by 2
$(8x^2+10x+2)(8x^2+10x+3)=90$
we can figure out that $8x^2+10x+2=9$ and $8x^2+10x+3=10$
$8x^2+10x-7=0$
$x_1+x_2=\frac{-b}{a}=\frac{-10}{8}=\frac{-5}{4}$
