SOLUTION: please help me solve. 1) (2x-3)(x-4) for an answer I got 2x^2-11x+12=0 is this correct 2) x^2+3x-28=2 3) x^2+16=-10x for an answer i got (x+2)(x+8)=0 is this one corr

Question 166490: please help me solve.
1) (2x-3)(x-4)
for an answer I got 2x^2-11x+12=0
is this correct
2) x^2+3x-28=2
3) x^2+16=-10x
for an answer i got (x+2)(x+8)=0
is this one correct
4) 5x^2=6x-16x^2
Answer by chiefman(11) (Show Source): You can put this solution on YOUR website!
1.(2x-3)(x-4)expading this yields
2x^2-8x-3x+12 adding like terms
2x^2-11x+12=0 and you dont have to worry since you were right.
2. x^2+3x-28=2putting likes together we have
x^2+3x-30=0 since this is not a perfect square use the quadratic formula

using the above formula we have
x=4.18 or 7.17
3.x^2+16=-10x putting -10x in the L.H.S we result to
x^2+10x+16=0 solving we get
(x+2)(x+8)=0 hence
x=-2 or -8
4.5x^2=6x-16x^2 grouping like terms
5x^2+16x^2-6x=0
21x^2-6x=0 getting the common term x
x(21x-6)=0
x=0,21x-6=0
21x=6
x=6/21 we therefore have
x=0 or 6/21
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