Will someone attempt to solve this problem for me?  This is a Quadratic Equation 15x squared + 28x + 5 = 0.?
969851
Basically there're 2 approaches to this problem:\n\n1. Factoring -- easy but NOT applicable to ALL problems.\nAfter factoring, it becomes:\n(5x+1)(3x+5)=0 (cuz 15=5*3, 5=1*5, and 28=5*5+3*1)\ntherefore either 5x+1=0 or 3x+5=0\nthen the two real roots are:\nx=-1/5, -5/3\n\n2. Quadratic formula -- general & reliable but more work. \nAssuming u know the formula (if not look it up in ur math book), just go head and plug in all corresponding coefficients of the original equation (here a=15, b=28, c=5), yields:\nx=(-28+/-square root of (28^2-4*15*5))/(2*15)=-1/5, -5/3\nwhich gives u exactly the same answers as method 1.\n\nHope these make sense to u.. GoOd Luck!..^^
447170
Change form to: \n\n5x^2 + 7x + 2 = 0 \n\nand use the quadratic formula. Or plot the graph of: \n\ny = 5x^2 + 7x + 2 \n\nand find the x-axis intersects. \n\nOr just do this: \n\n5x^2 + 7x + 2 = 0 \n\n5x^2 + 5x + 2x + 2 = 0 \n\n5x(x + 1) + 2(x + 1) = 0 \n\n(x + 1)(5x + 2) = 0 \n\nHence, \n\nx(1) = - 1 \n\nand \n\nx(2) = (-2)/5 \n\nsince one of the brackets has to equal zero to equal the left hand side. \n\nHope it's understandable and that it helped.
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a
1172065
if it is a quadratic (ax^2 + bx + c) then put it into the quadratic equation. that automatically will give you 0, 1, or 2 real answers. the equation is: x = [-b +/- (square root of b^2 - 4*a*c)] / 2a\n\nif (b^2 - 4*a*c) is negative, you have no real solutions. if it is zero, you have one real solution. if it is positive, you have 2 real solutions.\n\nthen, if for example you get  x=4 and x= -2, your factors are (x-4) and (x+2)
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
802559
You are given the quadratic function\ny = x² - 8x + 22\n\nTo find the vertex, change the given equation into the form\ny = a(x - h)² + k\n\nWhere \na = is a constant\n(h,k) is the vertex\n\nThus,\ny = x² - 8x + 22\n\nWe complete the square, by squaring the half of the coefficient of x (which is -8, the half of -8 is -4, and the square of -4 is 16.)  Thus,\ny = x² - 8x + 16 + 6\n\nNow, we can factor\ny = (x - 4)² + 6\n\nTherefore, the vertex is\n(4,6)\n\n^_^
838654
I would factor each quadratic and then see if I could somehow cancel some stuff out.\n\nx² - x - 6 = (x-3)(x+2)\ny² + 3y - 10 = (y+5)(y-2)\ny² - 7y + 10 = (y+5)(y+2)\n3x² + x - 10 = (3x-5)(x+2)\nx² - 2x - 3 = (x-3)(x+1)\n9x² - 25 = (3x+5)(3x-5)\n\nRewrite your problem with these factors and you will see that you can cancel out a lot of these terms!
583745
Try factoring each one.\n5x^3 - 15x^2 = 0\n5x^2(x-3) = 0 means\n5x^2 = 0 or x-3 = 0\nx = 0 or x = 3\n\nx^3 + 2x^2 - x = 2\nx^3 + 2x^2 - x - 2 = 0\nFactor by grouping.\n(x^3 + 2x^2) + (-x - 2) = 0\nx^2(x+2) + (-1)(x+2) = 0\n(x^2-1)(x+2) = 0\n(x-1)(x+1)(x+2) = 0\nx = 1, -1, or -2\nNow try the others on your own.
44408
5x2 = 9 − x\n\n5x2 + x − 9 = 0\n\nIt is a quadratic equation with\na = 5, b = 1, c = -9
528657
x^4 - 8x^2 + 15 = 0\n(x^2 - 5)(x^2 - 3) = 0\nSo x^2 = 5 or x^2 = 3.\nSo x = sqrt(5) or x = -sqrt(5) or x = sqrt(3) or x = -sqrt(3).\n\nTo purdue_engineer:\nIf it makes you feel any better, I've been screwed a bit lately.  One person copied my answer and he ended up getting best answer.  Another person was wrong but bashed my answer and so she got best answer instead of me!\n\nBut anyway, sorry!  I'm a teacher, and I just can't help myself.
939068
Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
598425
Solving for x in the first equation gives us...\n\nx + 5 - 5 < 4 - 5\nx < -1\n\nSolving for x in the second gives us...\n\n2 - 7x - 2 < 16 - 2\n-7x < 14\n---   ---   Here, don't forget to swap the sign\n -7    -7   since we are dividing by a negative.\nx > -2\n\nSo use that to find your solution set.  We know x must be less than -1 but greater than -2.  \n\nHope this is helpful.
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
1657
20x^2 + 15x - 8x -6 = 20x^2 + 7x - 6
922439
x = 15\n\n10 - 15 = -5
972476
........._______________\nx - 5 ) x³ - 3x² - 13x + 15 ( x²\n......... x³ - 5x²\n......... -------- ....↓\n............... 2x² - 13x ....... ( 2x\n............... 2x² - 10x\n............... ----------- ....↓ \n............... ....... -3x + 15. (-3\n............... ....... -3x + 15\n............... ....... -----------\n............... ....... -----------\n\nSo P(x) = x³ - 3x² - 13x + 15\n= (x - 5)(x² + 2x - 3)