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| Question 408261:  hi i need help with quadratic equations and i need steps please
 1)(x-6)(x-4)=0
 2)(x+2)(x-2)=0
 3)(x+6)(x+3)=0
 4)(x-4)(x+7)+0
 5)x(-7)=0
 6)(x-5)(x-2)=0
 7)(x-8)(x-6)=0
 8)(x+4)(x-4)=0
 9)x(x+5)=0
 10)(x-1)(x-8=0
 11)(x-2)(2x-1)=0
 12)(3x-2)(x+3)=
 13)(x+5)(4x-1)=0
 14)(2x-1)(3x+1)=0
 15)4x(2x-3)=0
 i need steps to please
 Answer by MathLover1(20850)
      (Show Source): 
You can put this solution on YOUR website! 1)(x-6)(x-4)=0
 the first one inside the brackets (x-6)=0 , and the second one inside the brackets (x-4)=0; the first one equals x=6, the second one equals x=4
 2)(x+2)(x-2)=0
 the first one inside the brackets (x+2)=0 , and the second one inside the brackets (x-2)=0; the first one equals x=-2, the second one equals x=2
 3)(x+6)(x+3)=0
 the first one inside the brackets (x+6)=0 , and the second one inside the brackets (x+3)=0; the first one equals x=-6, the second one equals x=-3
 
 4)(x-4)(x+7)=0
 the first one inside the brackets (x-4)=0 and the second one inside the brackets
 (x+3)=0; the first one equals x=4, the second one equals x=-3
 
 5)x(-7)=0
 since (-7) is not equal to zero, then x must be zero; x=0
 
 6)(x-5)(x-2)=0
 the first one inside the brackets (x-5)=0 and the second one inside the brackets
 (x-2)=0; the first one equals x=5, the second one equals x=2
 
 7)(x-8)(x-6)=0
 the first one inside the brackets (x-8)=0 and the second one inside the brackets
 (x-6)=0; the first one equals x=8, the second one equals x=6
 
 8)(x+4)(x-4)=0
 the first one inside the brackets (x+4)=0 and the second one inside the brackets
 (x-4)=0; the first one equals x=-4, the second one equals x=4
 
 
 9)x(x+5)=0
 the first one x=0, second one (x+5)=0; therefore second one x=-5
 
 10)(x-1)(x-8=0
 the first one inside the brackets (x-1)=0 and the second one inside the brackets
 (x-8)=0; the first one equals x=1, the second one equals x=8
 
 11)(x-2)(2x-1)=0
 the first one inside the brackets (x-2)=0 and the second one inside the brackets
 (2x-1)=0; the first one equals x=2, the second one equals 2x=1..therefore x=1/2
 
 12)(3x-2)(x+3)=0
 the first one inside the brackets (3x-2)=0 and the second one inside the brackets
 (x+3)=0; the first one equals 3x=2...or x=2/3, the second one equals  x=-3
 
 13)(x+5)(4x-1)=0
 the first one inside the brackets (x+5)=0 and the second one inside the brackets
 (4x-1)=0; the first one equals x=-5, the second one equals 4x=1..therefore x=1/4
 14)(2x-1)(3x+1)=0
 the first one inside the brackets (2x-1)=0 and the second one inside the brackets
 (3x+1)=0; the first one equals 2x=1 or x=1/2, the second one equals 3x=-1..therefore x=-1/3
 
 15)4x(2x-3)=0
 the first one 4x=0 only if x=0
 the second one inside the brackets
 (2x-3)=0;......2x=3 or x=3/2
 
 
 
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