(a(sq.) -4(sq.))Sq. - 4 (a(sq.)-4) - 32 =0?
931601
(a^2 - 4^2)^2 - 4(a^2 - 4) - 32 = 0\n(a^2 - 16)^2 - 4(a^2 - 4) - 32 = 0\na^4 - 32a^2 + 256 - 4a^2 + 16 - 32 = 0\na^4 - 36a^2 + 240 = 0\na^4 - 36a^2 = -240\na^4 - 36a^2 + 324 = 84\n(a^2 - 18) = +/- 84^(1/2)\na^2 = 18 +/- 84^(1/2)\na = +/- (18 +/- 84^(1/2))\n\nSo, there'll be four answers. Also, of course, x^(1/2) means the square root of x. Hope this helps, and good luck.\nP.S. Completing the square is always fun.
782446
x:-    0** 2** 4\ny:-  -4** 0** 4\n\ni hope that this helps
775585
-5+-sqr(25-4(2)(-9))/4\n-5+-sqr(97)/4\n1.21 and -3.71
540085
(a-2)(a+2)  \n(difference of squares)
347487
-6-9x=-9x-4\n-6=-4.....\nno solution?? whats so hard to that?
41284
Wait, all the answers are wrong\n\nx^2+4>4\nThen x^2-4x+4>0\nthen (x-2)^2>0\nBut hey, (x-2)^>0 is always true expect when x=2 which equals zero. An square is always positive or 0.\nSo the answer will be any value but 2 or R-{2}\nthese are the evidence:\nPut them in original non equation\nx=0 then 4>0 which is true\nx=-1 then 1+4>-4 so 5>-4 which is true\nx=3 then 9+4>12 so 13>12\n but at x=2 then 8>8 which is incorrect
124621
x-4y=-20\n=> -4y=-x-20\n=> 4y=x+20\n=> y=x/4+5 Ans.\nThanks.
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
345141
0 solutions!\n\nMove sqrt(x+8) on the other side and then raise by the power of 2. You will get\n\nx-4=4+4*sqrt(x+8)+x+8    (btw if there is a solution it has to be greater or equal to 4)\n\nthis is equivallent to \n\nsqrt(x+8)=-4\n\nWhich is impossible by the definition of sqrt(). It always has to be greater or equal to 0.
892842
Multiply the first equation by -2:\nx+4=-32\nSubtract 4:\nx=-36\nFor the second - multiply by -2/3:\n-14=x-2\nAdd 2:\nx=-12
592356
-(-3) X 4\n= 3 X 4\n= 12
764334
d=4\na=-1\ntn=a+(n-1)d\n=-1+(n-1)4\n=-1+4n-4\n=4n-5
861681
subtract 1st equation from the second:\n  8y^2=32\nso y^2=4, and y = +/- 2\n\nplug back into 1st equation:\n4x^2+4=4\n\n4x^2=0\nx=0
197379
x is anything less than 1.\n4 x1=4 so 1 cannot be the answer (unless the < had a line underneath it)\n4x0=0 which is less than 4\n4x -2= -8
325592
You need to first find 2 numbers that multiplies to give you a -4 and adds to give you a negative 3.  -4 and 1 work.\n(p^2 -4) (p^2+1) = 0\n\nThen, set them both = to 0.\np^2-4 = 0\np^2 = 4\np = 2\n\nand P^2 +1 = o\nP^2 = -1\nP = the square root of -1, also known as i\n\nFINAL ANSWER: P = 2, and i
169693
the absolute value of -4 (usually written |-4|) is 4\n\ntherefore (-2) - |-4| = (-2) - 4 = -6\n\n-6