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How do I solve this equation with the quadratic equation? 8(3x+5)^2+2(3x+5)-1=0

Ali242

by Ali242 at December 05, 2010

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You simplify it to start so it then reads:8(9x^2 + 30x + 25) + 6x + 4= 72x^2 + 246x + 204Then you plug in x = [-b±√(b^2 - 4ac)]/2aplugging it in you getx = [-246 ± √(492-58752)]/144x = undefined

yankeekid yankeekid December 05, 2010

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you can easily use substitution for this too. Let u = (3x+5)8u^2+2u-1=0(4u-1)(2u+1)=0now set u back to its original value(4(3x+5)-1) (2(3x+5)+1) = 0(12x+19) (6x+11) = 0x = [ -19/12 , -11/6 ]

fabianscorpio fabianscorpio December 06, 2010

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