solve  the following equation?
955903
3x^4-2x^2-8=0\nQuadratic formula with x^2 instead of x\n\na=3, b=-2, c=-8\n[2+-sqrt(4-4(3)(-8)]/6\n(2+-10)/6\nx^2=2 or x^2=-4/3\n\nx=+2 or -2\nx can also equal +sqrt(-4/3) or -sqrt(-4/3) but these are complex numbers
517123
It's hard to show it clearly here but the solution is actually:\n\n[-b±√(b²-4ac)]/2a\n\nthere will usually be two solutions!\nThe site below has it laid out more clearly...\n\n("invited" to leave school early.. 1974)
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
363348
Are we solving for x?
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).
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
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
943054
Distribute the 2 to the two terms in the parentheses. Group like terms. Solve for X.
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
580733
First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
1155111
xsquare=x+x+x...x terms\nby diff x\n2x=1+1+1...x terms\n2x=x\n2=1
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
901005
Use the following equation and solve:\n\nP1V1/T1=P2V2/T2\n\nT2=P2V2T1/P1V1\n\nalso note the significant figures!!!!
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
617358
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.