How do I solve the formula #16t^2 - 12t - h = 0# for #t#?

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Answer 1 · 2018-05-26T23:39:07

#t=(3+-sqrt( 4h+9))/8#

#16t^2 - 12t - h = 0#

you would need to complete the square:

#ax^2+bx +c#

#a=1#

#c=(1/2b)^2#

So first move the #h# over:

#16t^2 - 12t = h#

now divide by 16 so #a=1#

#(16t^2 - 12t)/16 = h/16#

#t^2 - 3/4t = h/16#

now we are ready to complete the square:

#t^2 - 3/4t +c = h/16 +c#

#c=(1/2b)^2#

#c=(1/2*-3/4)^2 = 9/64#

#t^2 - 3/4t +9/64 = h/16 +9/64#

#(t-3/8)^2 = h/16 +9/64#

#sqrt((t-3/8)^2)=+-sqrt( h/16 +9/64)#

#t-3/8=+-sqrt( h/16 +9/64)#

#t=3/8+-sqrt( h/16 +9/64)#

#t=3/8+-sqrt( (4h)/64 +9/64)#

#t=3/8+-sqrt( (4h+9)/64)#

#t=3/8+-sqrt( (4h+9))/8#

#t=(3+-sqrt( 4h+9))/8#