x^2 - 3x + 5\n\n\n(x^2 means x squared)
136288
Hello.  First you have to determine if you are going to find 'real' answers.  use b^2 - 4*a*c, where b=-3, a=1, c=5 (from ax^2 + bx + c).  \n\nNow, b^2 - 4*a*c = 3^2 - (4*1*5) = 9-20=-11, hence, you will have imaginary roots in your answer.\n\nNote that sqrt(-1) = i, then,\n\nx^2 - 3x + 5 = 0\n(x - [3/2])^2 + 5 - (3/2)^2 = 0 ;(Completing the roots)\n(x - [3/2])^2 = -11/4\n(x - [3/2])^2 = (i^2) * 11/4 ; (note that i^2 = -1)\n\nx - 3/2 = +/- i * 0.5 * sqrt(11)\nx = (3/2) +/- i * 0.5 * sqrt(11)\n\n------------\nAlways remember to check if b^2 - 4*a*c > 0 or < 0 then you will know if there is imaginary i in the answer..
341116
x^2-(x-3)(x-5)=185\nx^2-x^2+8x-15=185\n8x+-15=185\nadding 15\n8x=200\nx=25m\nnew dimensions=22m*20m\nchoice D
389812
2x - 2 = x^2 - 4x + 3\nSet one side equal to 0\n0 = x^2 - 6x + 5\nFactor\n0 = (x - 5)(x - 1)\nSet each factor equal to 0\n0 = x - 5              0 = x - 1\nSolve\n5 = x                  1 = x
523705
(x+2)/(x²-2) = (2)/(3-x)\nCross multiply\n(x+2)(3-x) = 2(x^2-2)\nSo -x² - x + 6 = 2x² -4\nSo 0 = 3x² + x - 10\n\nLooking for two numbers whose sum = +1 (ie b) and product = 3 * -10 = 30 (ie a*c) \nThese numbers are +6 and - 5\nSo + x = + 6x - 5x.\nSo 3x² + 6x - 5x - 10 = 0\n\n3x(x + 2) - 5( x + 2) = 0\n(3x - 5)(x + 2) = 0\nThus x = -2 or 5/3
428705
x/(x+5)-5x/(x²-25)\nx/(x+5)-5x/{(x+5)(x-5)\n{x(x-5)-5x}/{x²-25}\n(x²-10x)/(x²-25)
183119
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
66837
easy if that is the question\n2x+3x=5\nsince they have the same variable\nadd them \n5x=5\nthen divide by 5(no. to reduce the variable with 1 as its coefficient) \nx=1\n\nlonger method\n\n2(x)+3(x)=5\nby distributive property we know that (2+3)x=2(x)+3(x)\nthen: \n(2+3)x=5\n(5)x=5 divide by 5 \nx=1
85988
move everything to one side...\n\nx^3+5x^2-x-5=0\n(x^2-1)(x+5)=0\n\nsolve each parenthesis\n\n1). x^2-1=0\nx^2=1\nx=+1 and -1\n\n2). x+5=0\nx=-5\n\nso the answers are 1, -1, -5
352559
You only can say\nRoot[x^2 - 25x + 40]\nType for x sq only x^2.\n\nTh
417398
x³y - 3x²y - 10xy --- Take x and y out of each term...\nxy(x² -3x - 10) --- Factor the x² - 3x - 10\nxy(x - 5)(x + 2)\n\n*Note: To show exponents, such as x², use the ^ key (such as x^2). This expression looks like you basiclly wrote:\n3xy - 6xy - 10xy
321912
The equation is set up for you to solve by completing the square.  I would not multiply (x-5)^2 out (it is just extra work!)\n(x-5)^2 - 6 = 6\nAdd 6 to each side\n(x-5)^2 = 12\nTake the sqaure root of each side (remember to add a plus or minus sign.  I will use (+/-) to stand for the plus or minus sign).\nx-5 = (+/-)sqrt(12)\nAdd 5 to each side\nx = (+/-)sqrt(12) + 5\nSimplify the sqrt(12)\nsqrt(12)\n= sqrt(4*3)\n= 2sqrt(3)\nAnswer: x = (+/-)2sqrt(3) + 5
540045
Is (x+3) over 4 only, or over 4-2 ?  It's hard to tell from your problem, so I am going to assume it's over 4-2 = 2 unless you let me/us know otherwise by adding a comment to this question.\n\nCross multiply to solve:\n\n5 * (x+3) = (4-2) * (5x-2)\n\n5x + 15 = 10x - 4\n\nSubtract 5x from both sides:\n\n15 = 5x - 4\n\nAdd 4 to both sides:\n\n19 = 5x\n\nDivide both sides by 5:\n\nx = 19/5
329344
3x + x - 5 = 2(x-2)\n4x - 5 = 2x - 4\n2x = 1\nx = 1/2\n\n2) 6a^3 b^2 c / [2abc]\n= 3a^2 b\n\n3) 8x^2 y z^3 / [12x y^2 z]\n= 2xz^2/[3y]
47386
(x+3)(x^2-x+4)/(x+3) common factor\n\nmultiply by x+3\n\nx^2-x+4
521782
1)2(x+5)=-3x\n2)-5 and 11
207055
you subtract like terms.  \n5x^2 - 9x^2 = -4x^2\n-3x - (-5x) = -3x+5x = 2x\n6 - (-3) = 6+3 = 9\n\nTherefore, the final answer is -4x^2 +2x +9
532981
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!