b^2 + 6b -16 = 0 solve?
917375
b^2 + 6b -16 = 0\n(b+8)(b-2)=0\nb+8=0\nb=-8\n\nb-2=0\nb=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
258820
4x^2 -3x + 2 = 0 \nD = b^2 - 4ac = 9 - 4*2*4= 9- 32 = - 23<0\nThere are no real solutions, but there are 2 complex solutions.
931431
Let the roots be a and b\na+b=4\nab=3\n(a-b)^2=(a+b)^2-4ab=16-12=4\na-b=2\n\na=3, b=1 or a=1, b=3\n\nthe equation has roots 3 &1\nSo the equation is: (x-3)(x-1)=0\n\ni.e. x^2-4x+3=0
670230
x^2+4x+Y^2-8Y+16=0\nX^2+4x+Y^2-8Y=-16\nX^2+4x+4+Y^2-8Y+16=-16+16+4\nX^2+4x+4+Y^2-8Y+16=4\n(X+2)^2+(Y-4)^2=4\nR=Square root 4\nR=2
810055
x^2 - 2x + y^2 + 2y + 2 = 16\n\nx^2 - 2x + 1  + y^2 + 2y + 1 = 16\n(x - 1)^2 + (y+1)^2 = 16\n\nCenter = (1, -1)\nRadius = sqrt(16) = 4
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)
744314
multiply everything by b and (b-3)  ....... then you will get\n\n2b+7b-9=0\n\n(2b+9)*(b-1)=0\n\nso b= -9/2 or 1
633469
Set h(t) = 26 and solve,\n\n26 = 11 +34t -16t^2\n\n16t^2 - 34t +15 = 0\n\nnow, factoring by parts:\n\n(16t^2 - 24t - 10t +15)\n\n8t(2t - 3) - 5(2t - 3)\n\n(8t - 5)(2t - 3) = 0\n\nSo, it passes the window at t= 5/8 and t = 3/2.
526107
You are only allowed to factor if the equation is equal to zero; your's is equal to 4.  You need to solve "x^3 - 6x^2 +9x - 4 = 0"  using synthetic division.
929526
4x^2 - 16 = 0\n4(x^2 - 4) = 0\nx^2 - 4 = 0\n(x - 2)(x + 2) = 0\n\nANS : x = 2 or -2
531417
(a+b)(b^6-ab^5+a^2b^4-a^3b^3+a^4b^2-a^5b+a^6)\nGood Night, and Good Luck!
676772
If you must factor, this is how it is done...\n\n2x^4 - 16x^2 = 0\nx^4 - 8x^2 = 0\nx^2(x^2 - 8) = 0\nx^2 = 0 and x^2 = 8\nx = 0 and x = +/-sqrt(8)\n\nThere...
879052
x^2 - 8x + 16\n\n\nthe person above me said it is -x^2 but that is incorrect because -x * -x = x^2
195280
In this polynomial: a=6, b=-2, c=-20.\nSub a, b, and c into this equation: \nx= [-b+ square root the whole thing that inside this bracket(b*b -4ac)]/2a\n\nx'= [-b - square root the whole thing that inside this bracket(b*b -4ac)]/2a
986257
If the equation is X^4 - 12x^2 -64 =0, there is.\n\nUse u = x^2, now your equation will look like this:\n\nu^2 - 12u - 64 =0\nFind u1 and u2 using the quadratic formula.\n\nThe solutions are u1 = 16, u2 = -4\n\nNow you go back to x, first deal with u1, then u2.  Because u = x^2 each u solution will give you two x solutions.\n\n\nu = x^2 = 16\nx = +/- SQRT(16)\nx1 = 4\nx2 = -4\n\nx2 = -4.  This is not possible with real numbers. so it will give you two imaginary numbers. Do the same as before\n\nx = +/- SQRT(-4) = +/- 2*SQRT(-1) = +/- 2i\n\nx3 = i\nx4 = -i
188457
you can put 3w^2-6w-24=0 and solve this quadratic equation:\nw= [6+-√(36+4*24*3)]/2*3 = [6+-√(324)]/6=(6+-18)/6\nso w= (6+18)/6=24/6=4\nor w= (6-18)/6= -12/6= -2\nNow you can factor 3w^2 - 6w - 24 = 3(w -4)(w--2)=\n3(w-4)(w+2)