5x2-2x+16=4x2+6x ,or x2-8x+16=0, or x2-2(x)(4)+(4)2=0, or (x-4)2=0, or (x-4)(x-4)=0, or x=4
5x2 + 15x = 0 5x(x + 3) = 0 therefore, x = 0 or x + 3 = 0 that is, x = 0 or x = -3
5x2 + 2x - 3 = 0 5x2 + 5x - 3x - 3 = 0 5x(x + 1) - 3(x + 1) = 0 (5x - 3)(x + 1) = 0 So 5x - 3 = 0 or x + 1 = 0 ie x = 3/5 or x = -1
well 5x2=10 and anything x 0=0 so x=0 because 10x0=0
6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
5x2-2x+16=4x2+6x ,or x2-8x+16=0, or x2-2(x)(4)+(4)2=0, or (x-4)2=0, or (x-4)(x-4)=0, or x=4
5x2 + 15x = 0 5x(x + 3) = 0 therefore, x = 0 or x + 3 = 0 that is, x = 0 or x = -3
5x2 + 2x - 3 = 0 5x2 + 5x - 3x - 3 = 0 5x(x + 1) - 3(x + 1) = 0 (5x - 3)(x + 1) = 0 So 5x - 3 = 0 or x + 1 = 0 ie x = 3/5 or x = -1
-5x2 + 5x + 60 = 0 Remove the common factor of 5 -x2 + x + 12 = 0 Where the coefficient of x2 is negative, it is best to make it positive by multiplying throughout by -1 x2 - x - 12 = 0 The above equation will factorise. (x - 4)(x + 3) = 0 ; therefore x - 4 = 0 when x = 4, or x + 3 = 0 when x = -3 The solutions (roots) of this equation are x = 4 and x = -3.
x = 2 or x = 2/5
well 5x2=10 and anything x 0=0 so x=0 because 10x0=0
5x2+7x = 12 5x2+7x-12 = 0 (5x+12)(x-1) = 0 x = -12/5 or x = 1
5x2-2x = 0 x(5x-2) = 0 Therefore: x = 0 or x = 2/5
5x2 + 20x = 0 5x(x + 4) = 0 x = 0 and -4
6x2-2x+36 = 5x2+10x 6x2-5x2-2x-10x+36 = 0 x2-12x+36 = 0 (x-6)(x-6) = 0 x = 6 or x = 6 It has two equal roots.
Two. x = + i sqrt(8.4) x = - i sqrt(8.4)
you would first do 10+5x-5=0, then 5x+5=0 and is factored by 5(x+1)=0. or if you want to know what x equals with, x=-1