Solve for b.     a = 1/2 bc2?
827919
a = 0.5bc^2\n\n2a = bc^2\n\n2a/c^2 = b
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.
425927
Solve for b and c for a.\nb = 4a/3\nc = 5a/3\nso\na * 4a/3 * 5a/3 = 1620\n\ntimes both by 3\n\n3a * 4a * 5a = 1620 * 3 = 4860\n\n60a^3 = 4860\n\na^3 = 4860/60 = 729\n\na = 9\n\nb = 4* 9/3 = 4 * 3 = 12\nc = 5 * 9/3 = 5 * 3 = 15\n\nB = 12\n\ngood luck!
952904
A= (B*h)/2\nA=(10*15)/2\nA-=75
482048
-b (+/-) sqrt(b^2 - 4ac)\n--------------------------------- = x\n2a\n\nUsing the equation "ax^2 + bx + c = 0" to find the values of a, b, and c.
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
741426
If z=a+ib, then we have:\n2a=a²+b²\na²-2a+b²=0\nThis is a quadratic equation in a, with solutions:\n\na=(2±√(4-4b²)/2\na=1±√(1-b²)\n\nNote that since both a and b are real numbers, this implies that 0≤a≤2 and -1≤b≤1.
767438
b1=(-b+ (b^2-4ac)^1/2 )/2a=(-8+(64-48)^1/2)/6=(-8+4)/6\n    =-4/6=-2/3\nb2=(-b- (b^2-4ac)^1/2 )/2a=(-8-4)/6=-12/6=-2
61956
try to follow\n\nx/2-10=a\na/2-4=b\nb/2-2=6\n\nsolve and substitue\nb=16\na=40\nx=100 --the original amount\n\nhope that helped
550482
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
617358
x = (-b +- sqrt(b^2-4ac))/(2a)\n\nThis equation solves the quadratic equation, where\n\nax^2 + bx + c = 0.
762985
(b-4)/(b-4)(b+4) simplifies to 1/(b+4)\nrestrict denominator...bcannot equal 4 or -4
779257
put x=6ab/(a+b)in the eq(x+3a)/(x-3a) + (x+3b)/(x-3b).\nthen you get(9ab+3a^2/3ab-3a^2)+(9ab+3b^2/3ab-3b^2)\nby bringing the common values out in numerator and denominator and cancelling the them we get\n(3b+a/b-a ) +(3a+b/a-b)\nso the answer is 2
498337
from the first equation, x=(-a-b)/2\n\nsubstituting x in the 2nd:\n((-a-b)/2+a)^2=((-a-b)/2+b)^2\n((a-b)/2)^2=((b-a)/2)^2\n((a-b)^2)/4=((b-a)^2)/4\n(a-b)^2=(b-a)^2\na^2-2ab+b^2=a^2-2ab+b^2\n\nI disagree that this is not possible if a=b:\n(a+b)(a-b)+2x(a-b)=0\n\nif a=b, then:\n2a(0)+2x(0)=0\n2b(0)+2x(0)=0\n0+0=0\n\nHow does that not work? I think you are trying to say you can't divide by zero, but it is possible under both cases.
893931
m = 2\n\nb = -1\nso just substitute               these values in your equation\n\ny = mx + b\nthen you'll have\n\ny = 2x -1
457981
a/b = c/d\n\nthen ad = bc\n\nso 24x = 6x9\n\nx = 54/24\n\nx = 9/4