What kind of solutions does ax2-bx+c=0 have if b2-4ac<0?
965250
Two complex solutions because x= -b + or - the square root of b2-4ac all divided by 2a (if that made any sense, I can't really write it) and since you'd have to have the square root of a negative number, it has to be complex.  And since it's plus or minus, you have two answers!
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
508006
x^2 + 4bx + 2b=0\n(x+2b)^2 - 4b^2 + 2b = 0\n(x+2b)^2 = 4b^2 - 2b\n(x+2b) = SQRT(4b^2 - 2b)\nx = [SQRT(4b^2 - 2b)]-2b\n\n3x^2 -bx +1=0\n3(x^2 - (b/3)x + (1/3)) = 0\nx^2 - (b/3)x + (1/3) = 0\n(x-(b/6))^2 -(b/6)^2 + (1/3) = 0\n(x-(b/6))^2 = (b/6)^2 - (1/3)\nx-(b/6) = SQRT[b^2/36 - 12/36]\nx-(b/6) = [SQRT{^2 - 12)] / 6\nx = {[SQRT{^2 - 12)]+b} / 6\n\nb^2x^2 - 2bx + c^2= 0\n(bx)^2 - 2(bx) + c^2 = 0\n(bx-1)^2 - 1 + c^2 = 0\n(bx-1)^2 = 1 - c^2\nbx-1 = SQRT(1-c^2)\nbx = SQRT(1-c^2)+1\nx = [SQRT(1-c^2)+1]/b
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
754366
You need to use the quadratic equation as it is not factorable into rational expressions.\nThe form of the equation is:\nax^2 + bx + c = 0 \nSo in your case,\na = -2, b = 6, c = -3\nand the solution is:\nx = [ -b ± √(b^2 - 4ac) ] / (2a)\n\nso substitute and reduce.\n\nx = [ -6 ± √(6^2 - 4(-2)(-3)) ] / (2(-2))\n\nx = [ -6 ± √(36 - 24) ] / (-4)\n\n= [ -6 ± √12 ] / (-4)\n= [ -6 ± 2√3 ] / (-4) \n= (3 ± √3)/2
112444
Any method of solving will give x=0 y=0 and z=0 !
231877
ax^2+bx+c=0\nx^2+3x-9=0\n\nformula:\nfirst root:\n(-b+sqrt(b^2-4ac))/2a==> (-3+sqrt(3^2+4*9))/2=1,85\nsecond root:\n(-b-sqrt(b^2-4ac))/2a==> (-3-sqrt(3^2+4*9))/2=4,85
922434
There is only one step. Take the quadratic formula and plug in the numbers.\n\nIf ax^2 + bx + c = 0,\nx = (-b +/- sqrt(b^2-4ac))/2a\n\nFor the equation you gave, a=1, b=19, c=88. Plug those numbers in and see what you get.\n\nIf you're confused by the +/-, it's shorthand for writing two things in one. There are usually two values of x that work, one with + and one with -.
764329
x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
744954
f(x) = ax^2 + bx + c\n\nf(0) = c = 3\n\nf(1) = a + b + c = 7\n--> a + b = 4\n\nf(-3) = 9a -3b +c = 9*f(1) = 63\n9a -3b + c = 63\n9a -3b = 60\n9a - 3(4-a) = 60\n9a -12 +3a = 60\n12a = 72\na = 6\nb = -2\n\nf(x) = 6x^2 -2x +3
1051369
Given a+b+c=0\na+b=-c\nCubing both sides, we get\n(a+b)^3=(-c)^3\na^3+3a^2 b+3ab^2+b^3=-c^3\na^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc\nTherefore a^3+b^3+c^3=3abc
798499
Given ax² + bx + c = 0, the quadratic formula tells us that:\n\nx = [ -b ± √(b²-4ac) ] ÷ 2a\n\nIn your problem, a=3, b=12, c=1  --  simply plug these values into the quadratic formula to solve for x.
986056
the above is maximum when ax^2+bx+c is minimum\n\nax^2+bx+c is minium or c when x = 0 as a,b are positive\n\nso maximum value f(x) is at 0 that is A exp(-c)
954093
k1ax - k2x^2 = 0\nx(k1a - k2x) = 0\n\nat x = 0 and...\nk1a - k2x = 0\nx = k1a/k2\n\nso at x = 0 and a*k1/k2
940135
Use factoring.\nx2-2x-8=0\n(x-4)(x+2)=0\nso\nx-4=0 and x+2=0\nsolve\nx=4 and x=-2
293371
The equation has 0 real solutions, but 2 complex solutions.\n\nThere is a very fundamental theorem of algebra which states that any polynomial equation of degree n has n roots. In your case, the degree is 2, so the equation has to have 2 roots.\n\nHowever, roots can be either real, complex, or repeated roots. For example, x^2=0 appears to only have one solutions (x=0) but x=0 is a repeated root, occurring twice. For polynomials of degree 2, you can find out about the solutions by using the discriminant. An equation of the form ax^2+bx+c=0 has discriminant b^2-4ac. The discriminant can be one of three things:\n\n>0 This means that both roots are REAL roots.\n=0 This means that there is ONE REAL root, and it is a repeated root.\n<0 This means that both roots are COMPLEX roots and there is no real solution.\n\nThe methods for general degree n polynomials are much more difficult, if not impossible, to classify.
296877
It has a solution, indeed. It actually has two solutions, -3 and 7. To find it out, set each pair of nomials, (e.g. (x + 3)) to zero. So it would be (x+3)=0. Then solve for x. Do the same for the other nomial, and there you have it.