Question 974325
Rewrite first as 2x^2-3x+7=0

x=(1/4)[3 +/- sqrt(9-56) ;;;; sqrt(-47)

roots are (3/4) +/- i sqrt (47)

{{{graph(300,300,-10,15,-10,250,2x^2-3x+7,x^2+7x+12)}}}

Not clear if these are supposed to be simultaneous equations.

x^2+7x+12=0 is the second

x^2+7x+12=2x^2-3x+7
-x^2+10x+5=x^2-10x-5
That solution (common to both equations) is when x=(1/2)[10+/- sqrt(100+20)]=(1/2)[10 +/- 2 sqrt(30]

Equal at 5 +/-5.47;; -0.47 and 10.47 would be approximate x-solutions.
Is this what you wanted?

equal at x=0.55 and 9.45