Give the roots of the equation  5^2x -265^x +25 = 0?
828200
No.  Do your own f*cking homework.
988599
2^1999 x 5^2000 = \n2^1999 x 5^1999 x 5 =\n(2 x 5)^1999 x 5 =\n5 x 10^1999\nThis is 5 followed by 1999 zeroes.\nThis an exact aresult.
512448
x^2 + y^2 = 25\nx - y = 0\n\nx = y\n\nY^2 + Y^2 = 25\n2 Y^2 = 25\n\nY ^2 = 25/2\ny =  (25/2)^1/2 or\ny = Sqrt (12.5)
255534
5^2 = 25\n100^2 = 10,000\n15^2 = 225\n\nso 25 + (10,000*225) = 2,250,025\n\nor did you mean\n\n(25 + 10,000) * 225 = 2,255,625
927994
* Start with the identity\n          − 20 = − 20\n\n    * Express both sides in slightly different, yet equivalent ways\n          25 − 45 = 16 − 36\n\n    * Factor both sides\n          5^2 - 5 &#92;times 9 = 4^2 - 4 &#92;times 9\n\n    * Add the same thing to both sides\n          5^2 - 5 &#92;times 9 + &#92;frac{81}{4} = 4^2 - 4 &#92;times 9 + &#92;frac{81}{4}\n\n    * Now factor both sides again\n          &#92;left(5 - &#92;frac{9}{2}&#92;right)^2 = &#92;left(4 - &#92;frac{9}{2}&#92;right)^2\n\n    * Take the square root of both sides\n          5 - &#92;frac{9}{2} = 4 - &#92;frac{9}{2}\n\n    * Cancel the common term\n          5 = 4\n\n    *Proof \n           2+2=5
282386
First, I suggest that when you encounter the words "square root" you immediately translate to "the one half power." Now your equation is f(x)= 5*[(5x - 5)^(1/2)] -6; write that as 5*[(5x - 5)^(1/2)] -6 = y, so that f(x) = y. Now exchange the variables, x for y, and y for x. The new equation becomes 5*[(5y - 5)^(1/2)] -6 = x. Solve this equation for y and you now have the equation for the inverse of the function f(x), specifically f'(x). Step by step: Add 6 to each side, then divide both sides by 5; you get [(5y - 5)^(1/2)] = (x+6)/5. Now square both sides: (5y - 5) = [(x+6)/5]^2. Add 5 to each side: 5y = [(x+6)/5]^2 + 5; then divide both sides by 5: y = {[(x+6)/5]^2 + 5}/5 = {[(x+6)/5]^2 }/ 5 + 1. Further manipulation if you wish, yields  {[(x+6)^2]/25}/5 + 1, and a final form of [(x+6)^2/125] + 1 = y = f'(x), the inverse function. All you do is exchange the x's and y's and solve for y in the "exchanged" equation.
706256
x(rt2)x+5rt2(x^2)-25y(rty)\n=6rt2x^2-25yrty
271143
Is that 2 times n cubed?  If it is, the answer is n = 5.\n\n250 = 2*n^3\nn^3 = 250/2  = 125\nn = cube root of 125 = 5\n\nto check, 5 x 5 x 5 = 25 x 5 = 125
1073032
Urs is wrong.Here is the correct answer.------>\nGiven equation, 16 x^2 +25 y^2= 100\n\n=>  x^2/(100/16) +y^2/(100/25)=1\nWe know,\ne= square root over[ 1-b^2/a^2]\n\nso, here b=100/25 and a=100/16\nTherefore, e=sq root over[ 1-(16/25)]\n after simplification e=3/5 or 0.6.
952526
x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
1067708
2x^2 -5x - 6 =0\n\nthe roots\nr1 = (- b + SQRT (b^2 - 4ac) )/2a \nr2 = (- b -  SQRT (b^2 - 4ac) )/2a \nwhere\na = 2  , b= -5 , c = -6\n\nso r1=(5+ sqrt (25+48)/4) = 3.386\nr2 = (5-sqrt (25+48)/4) = -0.886\n\nBy the way, the answer just above mine has an error, the √-13 should be √73 cos it's actually √ 5^2 - 4 (2)(-6) = √25 + 48 = √73\n\n_____________________________________________\nAGAIN, YOU JUST NEED TO CHANGE THE COEFFICIENT C\n-------------------------------------------------------------------------\nafter your details added , you want to solve\n2x^2 -5x - 3 =0\n\nthe roots\nr1 = (- b + SQRT (b^2 - 4ac) )/2a \nr2 = (- b -  SQRT (b^2 - 4ac) )/2a \nwhere\na = 2  , b= -5 , c = -3\n\nso r1=(5+ sqrt (25-(4)(2)(-3))/4) = (5+sqrt(49))/4 = 12/4 = 3\nr2 = (5-sqrt (25-(4)(2)(-3))/4) = (5-sqrt(49))/4 = -2/4 = -1/2
1108240
2(x-5)^2 = 3\ndivide by 2:  (x - 5)^2 = 3/2\nsquare root: (x - 5) = +/- root(6)/2\nadd 5: x = 5 +/- root(6)/2
629654
x^2+y^2-2y-5=0\n\nx^2 + y^2-2y + 1 - 1 -5 =0\n(x-0)^2 + (y-1)^2 = 6\n\nCenter (0, 1) radius sqrt(6)
921698
let the base be x\narea=1/2*x*[5^2-(x/2)^2]\n       =1/2x[25-x^2/4]\ndA/dx=1/2[x(-1/2x)+(25-x^2/4)]\nsetting it to zero\n1/4x^2=-x^2/4+25\nx^2/2=25\nx^2=50/3\nx=5rt(2/3)\nh=rt(25-50/12)\n=rt250/12\n=5/2rt(10/3)\narea=(1/2)5rt(2/3)(5/2)rt(10/3)\n=25/4rt(20/9)\n=25/12rt20\nx=10/rt3\narea=1/2*10/rt3*200/12
849035
1/2\n\nIf z^5 = 1, then z^5 - 1 = 0\nor, after factoring,\n(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0\n\nIf a and b are roots of this equation, and neither a nor b is =1, then a and b both satisfy\nz^4 + z^3 + z^2 + z + 1 = 0\n\nso a^4 + a^3 + a^2 + a + 1 = 0\nand b^4 + b^3 + b^2 + b + 1 = 0\n\nso 2+a+a^2+a^3+a^4 = a^4 + a^3 + a^2 + a + 1 + 1 = 0+1 = 1\nand 3+b+b^2+b^3+b^4 = b^4 + b^3 + b^2 + b + 1 + 2 = 0+2 = 2\ntherefore\n(2+a+a^2+a^3+a^4) / (3+b+b^2+b^3+b^4) = 1/2
803969
x^2+2x-35+39=39\nx^2+2x+4=39\n(x+2)^2 = 39\nx+2 = sqrt 39\nx = (sqrt 39) - 2 \n\nx^2-4x-2=0\nx^2-4x-2+6=6\nx^2-4x+4=6\n(x-2)^2 = 6\nx-2 = sqrt 6\nx = (sqrt 6) + 2
193529
if your equation is of the form\nax^2 +bx+c=0 then the general formula for the solution is\nthe two roots are:\n(-b+(b^2-4ac)^(1/2))/(2a)\n(-b-(b^2-4ac)^(1/2))/(2a)\nif b^2-4ac is less than zero then you will have imaginary roots, in such a case ur roots would be\n(-b+i(4ac-b^2)^(1/2))/(2a)\n(-b-i(4ac-b^2)^(1/2))/(2a)