Question 208967
I need to solve for t

<pre><font size = 4 color = "indigo"><b>
{{{0 = 3t^2 + 4t - 24}}}

{{{3t^2 + 4t - 24=0}}}

There are two ways, completing the square
and quadratic formula.

COMPLETING THE SQUARE:
Get all variable terms on the left and
the constant term on the right:

{{{3t^2 + 4t = 24}}}

Divide through by the coefficient of {{{t^2}}}

{{{(3/3)t^2+(4/3)t=24/3}}}

{{{  t^2+(4/3)t=8 }}}

Multiply coefficient of t by {{{1/2}}}:  {{{(4/3)(1/2)}}}={{{2/3}}}
Square that: {{{(2/3)^2=4/9}}}
Add 4/9 to both sides:

{{{t^2+(4/3)t+4/9=8+4/9}}}

Factor left side, get LCD on the right:

{{{(t+2/3)(t+2/3)=72/9+4/9}}}

{{{(t+2/3)^2=76/9}}}

Take square roots of both sides, remembering ± on the right

{{{sqrt((t+2/3)^2)=""+-sqrt(76/9)}}}

{{{t+2/3=""+-sqrt(76)/sqrt(9)}}}

{{{t+2/3=""+-sqrt(76)/3}}}

Add {{{-2/3}}} to both sides:

{{{t=-2/3+-sqrt(76)/3}}}

Break up 76 by observing that it can
be divided by perfect square 4, so
76=4*19

{{{t=-2/3+-sqrt(4*19)/3}}}

{{{t=-2/3+-sqrt(4)sqrt(19)/3}}}

{{{t=-2/3 +- 2sqrt(19)/3}}}

So the two solutions are

{{{t=-2/3+2sqrt(19)/3}}} and {{{t=-2/3-2sqrt(19)/3}}}


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QUADRATIC FORMULA

The solution to {{{ax^2+bx+c=0}}} is

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

with x=t, a=3, b=4, c=-24

{{{t = (-(4) +- sqrt( (-4)^2-4*(3)*(-24) ))/(2*(3)) }}}  

{{{t = (-4 +- sqrt(16+ 288 ))/6 }}} 

{{{t = (-4 +- sqrt(304 ))/6 }}}

{{{t = (-4 +- sqrt(16*19 ))/6 }}}

{{{t = (-4 +- sqrt(16)sqrt(19) )/6 }}}

{{{t = (-4 +- 4sqrt(19) )/6 }}}

Make two fractions:

{{{t = -4/6 +- 4sqrt(19) /6 }}}

Reduce the fractions by dividing top and bottom by 2:

{{{t = -2/3 +- 2sqrt(19) /3 }}}

So the two solutions are

{{{t=-2/3+2sqrt(19)/3}}} and {{{t=-2/3-2sqrt(19)/3}}}

Same answer.

Edwin</pre>