You fill in for x and solve the equation
3(5)^2 - 2(5)
15^2 - 10; now solve
3x2 - 2x - 5 = (3x - 5)(x + 1) = 0 gives a solution of x = 5/3 or -1. Was that what you wanted?
2(2x-11)(3x+5)
6 - 3x = 4 - 2x 6 = x + 4 x = 2 Check it. 6 - 6 = 4 - 4 It checks.
Let square = [x] -4+3*(-2)*[-4]+[3x-2x]=-4-6-2+[x]=-12+[x]The answer is negative 12 plus squared x.
The discriminant b2-4ac is -23
3x2 + 2x - 8 = 0 is a quadratic equation.
3x2 - 2x - 5 = (3x - 5)(x + 1) = 0 gives a solution of x = 5/3 or -1. Was that what you wanted?
x(3x - 2)
It is a quadratic equation and its solutions are: x = -3/2 and x = 3
3x=2x-12x=-12
3x^(2) +9x - 2x -6 Collect 'like terms'. Hence 3x^(2) + 7x - 6 Next write down all the factors of '3' and '6' Hence 3 ; 1' & 3' 6 ; 1,6 ; 2,3. From these pairs of number we select a pair from each coefficient, that add/multiply to '7' . Hence (3' x 3 ) & (1' x 2) ; NB 'dashes' (') to indicate source of numbers. Write up brackets (3x 2)(x 3) -2)(x + 3) Next we notice that the '6' is negative, so the two signs are different (+/-) The '7x' is positive , so the larger number takes the positive sign . Hence (3x - 2)(x + 3)
5x3 - 3x2 + 2x = x*(5x2 - 3x + 2) The quadratic has no real factors.
(x - 2)(2x + 7)
3X = x2 - 2xsubtract 3X from each sideX2 - 5X = 0factor out an XX(X - 5) = 0=========X = 0--------orX = 5--------
3x - 2x + 4 = 0 3x - 2x = -4 x = -4
3x(2x + 1)
3x+7=2x-5 x=-12