solve the following equation?
819492
3^4 = -2x^2 + 8\n81 = -2x^2 + 8   simplify\n73 = -2x^2          subtract 8 from both sides\n-73/2 = x^2         divide both sides by -2\nsqrt(73/2) * i = x   take square root of both sides... since the right is negative you get an imaginary number.  i = sqrt(-1)
753426
X2   5x - 4y = 1\n10x -8y =2\n-10x + 8y = -3\n-------------------------\n  0=-1                NO SOLUTION
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
589850
e)\nSolve the eqn.\nG(k)=k^2-0.5k=-0.5\nk^2-0.5k+0.5=0\nk= (0.5+_sqrt(.25-2))/2\n0.25+-sqrt(-1.75) /2\nSince term inside sqrt is negative \nthere are no real solutions
617282
are u sure its x2? shouldn't it be 2x+10x=(a number) usually there is a number at the end of the equation and u have 2 solve for "x". add details to ur question and then i may be able 2 help u.
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
605879
(x+3)(x+12)(x+2)(x+8)= [(x+3)(x+8)][(x+12)(x+2)]\n                     = (x^2+11x+24)*(x^2+14x+24)\nlet t=x^2+24+(25X)&#92;2 . \nSo we have: x^2+11x+24=t-(3x)&#92;2 and x^2+14x+24=t+(3x)&#92;2                                 \nso the equation becomes: [t-(3x)&#92;2][t+(3x)&#92;2] = 4x^2\n                    <-->t^2-(9x^2)&#92;4=4x^2  \n                    <-->t^2=(25x^2)&#92;4\n                    <-->t=(5x)&#92;2 or t=-(5x)&#92;2\n                    <-->x^2+10x+24=0 or x^2+15x+24=0\ncontinue by yourself
364111
take the first derivative and set it equal to zero and solve.  2x +1 = 0 x= -1/2  solve for y.  y=-13/4\nvertex is at x= -13/4 y=-2
617358
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.
984409
The goal here is to find the equation with x on one side and everything else on the other.  It's like the equation you have, but instead of y= you want x=.  So you need to "solve for x."  Think of y as just a number, and you need to find what x =.\n\nStep by step:\n\ny=4x+4\n\nSubtract 4 from each side:\n\ny-4=4x\n\nDivide both sides by 4\n\n(y-4)/4=x\n\nx=(y-4)/4\n\nThere's your equation!  Hope this helps.
784758
x^2+x+1=0\nx^2+2(1/2)x+1=0\nX^2+2(1/2)x+(1/2)^2+1-(1/2)^2=0\n(x+1/2)^2+3/4=0\n(x+1/2)^2=-3/4\nx=-1/2+/-√(3/4) i=-1/2+/-√3 * i/2
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2