SOLUTION: please help me solve 1) 6x^2+24x=2x^2-36 2) 3x^2-17x+20=0 for this one i got (x-6)(3x-5) is this correct 3) x^2-36=0

Question 166491: please help me solve
1) 6x^2+24x=2x^2-36
2) 3x^2-17x+20=0
for this one i got (x-6)(3x-5)
is this correct
3) x^2-36=0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1) 6x^2+24x=2x^2-36
4x^2 +24x + 36 = 0
x^2 + 6x + 9 = 0
(x+3)^2 = 0
x = -3 with multiplicity 2
===================================
2) 3x^2-17x+20=0
3x^2 - 12x - -5x + 20 = 0
3x(x-4)-5(x-4) = 0
(x-4)(3x-5) = 0
x = 4 or x = 5/3
=====================
3) x^2-36=0
(x-6)(x+6) = 0
x = 6 or x = -6
======================
Cheers,
Stan H.

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