Ok need help with quadratic equasion.?
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 -.
756992
Don't you just love the quadratic equation?\n\nThe first thing you got to do is get the given equation into the correct form, and in order to do that you have to combine all like terms. Since there are no like terms yet you have to do some math first.\n\n-2.0^2 = 4.0because neg * neg = pos\n\nNow you have 2 like terms 4.0 and 0.5 combine to get 4.5. Your eqution should look like this:\n\nf(t)=10t + 4.5 since you need at^2 + bt^2 + c just insert t^2. Now your equation will look like this:\n\nf(t)=t^2 + 10t + 4.5\n\nAnd just in case you need the quadratic equation:\n\nt= -b + or - sqr(b^2 - 4ac) / 2a\n\noh yeah and a = 1
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
82879
If your quadric equation is ax2 + bx + C = 0 Than your program should be as below to find the value/s of X.\n\n#include<stdio.h>\n#include<conio.h>\n#include<math.h>\n\nvoid main()\n{\n\nclrscr();\nfloat a,b,c,delta,x1,x2,sqrtdelta;\n\nprintf("Enter the Value of a : ");\nscanf("%f",&a);\nprintf("Enter the Value of b : ");\nscanf("%f",&b);\nprintf("Enter the Value of c : ");\nscanf("%f",&c);\n\ndelta= b*b - 4*a*c;\nsqrtdelta = sqrt(delta);\n\nx1 = (-b + sqrtdelta)/(2*a);\nx2 = (-b - sqrtdelta)/(2*a);\n\nprintf("The X1 : %f",x1);\nprintf("The X2 : %f",x2);\n\ngetch();\n\n}\n\nI think this enough for you. If you need any further help than try to contact me.\n\nBye.
530142
use the quadratic equation:\nx=-b +/- the square root of b^2 - 4ac all over 2a\na=1,b=7,c=10 so plug it in\nx=-7 +/- the square root of 49-40 all over 2\n-7 +3 = -4 divided by 2 x=-2\n-7-3=-10 divided by 2 = -5\nso x = -2 or x=-5
826907
ill write down the formula for finding the root of a quadratic equation :  ax^2 + bx + c = 0\nsolution: x = {-b +- sqrt(b^2 - 4ac) } / 2a\n\nequation x^2=5x + 2 can be written as :\n\nx^2 - 5x - 2 = 0\nhere a = 1 ; b = -5 ; c = -2 ;\nsimply apply the formula to find the value of x .\n\nx ={ -(-5)+-sqrt(25 - 4*-2) } /  2\nx ={ 5 +- sqrt(24+8) } / 2\nx = { 5 +- sqrt(33) } / 2\n\nThe second equation :\n\n4x^2 + 2x - 3 = 0\nhere a = 4 ; b = 2 ; c = -3\n\nx = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4\nx = {-2 +- sqrt(4+48) } / 8\nx = { -2 +- sqrt(52) } /8\n\nsimplify the term inside the square root ,  52 = 13*4\nso 4 gets square rooted and\n\nx = { -2 +- 2 sqrt(13) } /8\n\ntake 2 common from the numerator and denominator and cancel them.\n\nx = 2{ -1 +- sqrt(13) } / 4*2\nx = { -1 +- sqrt(13) } / 4\n\nthats your answer.\n\nhave a nice day :)
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~ ♥ ~
154468
All you have to do for substitution is substitute one equasion into another so what you do is:\n1. x+3(x+3)=1 because y=x+3\n2. multiply it out to get x2(squared)+6x+9=1\n3. subtract the one so that the equasion equals 0 x2(squared)+6x+8=0\n4. factor the equasion to get (x+2)(x+4)=0\n5. the answer for x could be -2 or -4 because either one of those would make one of the () 0 which would make the equasion equal 0\n6. becasue x=-2 or -4 y=1 or -1\n7. check the answers
932849
The quadratic formula goes: (-b ± √b² - 4ac)/ 2a\nSo x²=a, -6x=b, -7=c\nNow plug it in:\n [-(-6) ± √(6)² -4(1)(-7)]/ 2(1) \nsimplify: (6± √36 + 28)/ 2\nsimplify further: (6±√64)/2\nwhich equals: (6±8)/2\nand if you separate it out: (6-8)/2 = -1      and   (6+8)/2 = 7
857762
Not sure if it's the shortest or nicest way, but I was able to get to a solution using substitution u=x^2, and v=y^2, then adding 1/4 to the v side to make it a quadratic equation, solving for v, then re-substituting, and solving for y.\n\nI got y = sqrt ( sqrt (x^2-x^4+1/4) + 1/2 )\n\nYou should be able to plot that ...
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
953940
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
896553
4x^2+8x-x-2\n(4x-1)(x+2)
17048
= x^2 - 7x + 4x -28\n= x^2 - 3x - 28
1007589
1. b^2-4ac>0 : b>2sqrt(ac)\n2. b^2-4ac=0 : b=2sqrt(ac)\n3. b^2-4ac<0 : b<2sqrt(ac)\n\nThe rest should be a piece of cake.
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
705433
Derivation of the quadratic formula by completing the square.\n\nax^2 + bx + c = 0\nx^2 + (b/a)x + (c/a) = 0\nx^2 + (b/a)x = -c/a\nx^2 + (b/a)x + b^2/(4a^2) =  b^2/(4a^2) - c/a \n(x + b/2a)^2 =  b^2/(4a^2) - 4ac/(4a^2) \nx + b/2a = ±√[ b^2/(4a^2) - 4ac/(4a^2) ]\n\nx = -b/2a ± √[ b^2/(4a^2) - 4ac/(4a^2) ]\n= -b/2a ± √[ b^2- 4ac] / 2a\n\nx = {-b ± √[ b^2- 4ac] } / 2a