<s>[INST] 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!
 [/INST] $(x+1)(x+\frac14)(x+\frac12)(x+\frac34)=\frac{45}{32}$
$(x+1)(4x+1)(2x+1)(4x+3)=45$
$(4x^2+5x+1)(8x^2+10x+3)=45$
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 $8x^2+10x+3=10$
$8x^2+10x-7=0$
$x_1+x_2=\frac{-b}{a}=\frac{-10}{8}=\frac{-5}{4}$
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