Solve by completing the sqaure x2-2x-5=0?
954896
x^2 + -2x = 5   Isolate the x^2 and x terms.\n\nx^2 + -2x = 5   Divide both sides of the equation by 1, which is the a-coefficient.\n\nx^2 + -2x + (-2/2)^2 = 5 + (-2/2)^2   Add half of the b-coefficient squared. In this equation that value is (-2/2)^2, and can be simplified to 1.\n\n(x + -2/2)^2 = 5 + (-2/2)^2   Write the x^2 and x terms side as a perfect square binomial.\n\n(x + -2/2)^2 = 5 + 1   Simplify the other side (simplify fractions).\n\n(x + -2/2)^2 = 6   Simplify the other side (combine numbers/fractions).\n\nSQRT[(x + -2/2)^2] = {+,-}SQRT[6]   Find the square root of both sides.\n\nWhen you combine and solve you'll get\n\nx = -1 plus or minus SQRT[6]
44408
5x2 = 9 − x\n\n5x2 + x − 9 = 0\n\nIt is a quadratic equation with\na = 5, b = 1, c = -9
769640
Solve one equation for x (or y), then substitute that value into the other equation.\n\nExample:  x-2y=5, so x=2y+5\n\nNow substitute this for x in "5x-2y=5"\n\n5(2y+5) - 2y = 5\n\n10y + 25 - 2y = 5\n\n8y = -20\n\ny = -20/8 = -5/2\n\nPut this value for y back into your first equation to solve for x:\n\nx = 2y + 5\nx = 2(-5/2) + 5\nx = -5+5\nx = 0\n\nANSWER: x=0, y=-5/2
781653
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
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-5+-sqr(25-4(2)(-9))/4\n-5+-sqr(97)/4\n1.21 and -3.71
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(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
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2(x - 5)^2 = 3\nSo (x - 5)^2 = 3/2\nThus x - 5 = ± √(3/2)\n= ± √6 / 4\n\nThus x = 5 ± √6 / 4
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assuming your expression equals 0\nx^2-25-0\n(x+5)(x-5)=0\nx=5 or -5
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I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
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I hope you understand Algebra...\n\nx² - 5x - 4 = 0\n\nTranspose -4\nx² - 5x = 4\n\nAdd 25/4\nx² - 5x + 25/4 = 4 + 25/4\n\nFactor the perfect square trinomial and change 4 into 16/4\n(x - 5/2)² = 16/4 + 25/4\n\nCombine like terms\n(x - 5/2)² = 41/4\n\nget the square root of both sides\nx - 5/2 = (sqrt41)/2\n\ntranspose -5/2\nx = 5/2 + (sqrt41)/2\n\nFuse them into one expression\nx = (5 +or- sqrt41)/2\n\nThere are 2 possible answers\nx = (5 + sqrt41)/2\nor\nx = (5 - sqrt41)/2\n\nthis Method is called "completing the square"\n\nHope this helps^_^
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Completing the square involves creating a perfect square trinomial.  Consider\n\nx^2 + 4x - 1 = 0\n\nCompleting the square would involve trying to make one side a perfect square: (x + 2)^2 in this case.  Step by step:\n\n1.  Get all your x terms on one side, and constants on the other.\n\nx^2 + 4x = 1\n\n2.  Take half of the coefficient of the x term and square it.  Half of 4 is 2, and 2 squared is 4.\n\n3.  Add this to both sides of the equation.\n\nx^2 + 4x +4 = 1 + 4\n\n4.  Now factor the left side.  It works out to be a perfect square.\n\n(x + 2)^2 = 5\n\n5.  Take the square root of both sides and solve for x.\n\nx + 2 = +/- sqrt(5)\n\nx = +/- sqrt(5) - 2\n\nThe quadratic equation is pretty easy:\n\nx = [-b +/- sqrt (b^2 - 4ac)] / 2a\n\nIn the above example, a = 1, b = 4, and c = -1 (coefficients of the x^2 and x terms are a and b, and the constant is c).  So plugging in to the quadratic formula gives you\n\nx = [-4 +/- sqrt(4^2 - 4x1x(-1))] / 2x1\n  = [-4 +/- sqrt(16 - (-4))] / 2\n  = [-4 +/- sqrt(20)] / 2\n  = [-4 +/- 2sqrt(5)] / 2\n  = -2 +/- sqrt(5)\n\nwhich is the same as we got via completing the square.
19885
(5) x (2x^2-3x+5) I think
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You have to solve the equation so that you can eliminate the sqrt by raising it to the second power .for instance,\n\nsqrt(2x+10)=5     you square it and get\n\n2x+10=25   easy, isnt it?\n\nbe careful if you have something like this\n\nsqrt(x+3)+sqrt(x-2)= 5\n\nhere you cannot do that you have to distribute one sqrt on each side of the equation.\n\nsqrt(x+3)=5-sqrt(x-2)\n\nnow you can square them
180232
since x=0, y=2*0+5=5\nthe answer: y=5.
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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
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not sure what box method is...the answer is\n(2x+5)*(x-3)
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x={-(-4) (+/-)√[(-4)^2 - 4(1)(-5)]}/(2(1))\n x= 4(+-)6/2\n x=5,   x=-1\n\nCheck:\n with x=5,  5^2-4(5)-5=25-20-5=0  OK\n with x=-1   (-1)^2 -4(-1) -5 = 1+4-5=0  OK