Question 181102
complete the square?
1.) 3t^2 + 24t-13
The coefficient of t^2 has to be 1, divide each term by 3, results:
t^2 + 8t - {{{13/3}}} = 0
:
t^2 + 8t + ___ =  {{{13/3}}}
Find the term that will complete the square; square half of the coefficient of t
.5*8=4; 4^2=16; add to both sides
t^2 + 8t + 16 =  {{{13/3}}} + 16
:
t^2 + 8t + 9 =  {{{13/3}}} + {{{48/3}}}; change 16 to thirds
:
(t + 4)^2 =  {{{61/3}}}
Find the square root of both sides:
t + 4 = +/-{{{sqrt(61/3)}}}
subtract 4 from both sides, we have two solutions
t = -4 +{{{sqrt(61/3)}}}
and
t = -4 -{{{sqrt(61/3)}}}
:
:
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2.) 4s^2+s+2
Divide each term by 4 to make the coefficient of s^2 = 1, results
s^2 + {{{1/4}}}s + {{{2/4}}} = 0
s^2 + {{{1/4}}}s + ___ = -{{{2/4}}} 
Complete the square, half of (1/4) = (1/8); (1/8)^2 = (1/64)
s^2 + {{{1/4}}}s + {{{1/64}}} = -{{{2/4}}} + {{{1/64}}}
s^2 + {{{1/4}}}s + {{{1/64}}} = -{{{32/64}}} + {{{1/64}}}
s^2 + {{{1/4}}}s + {{{1/64}}} = -{{{31/64}}}
(s + {{{1/8}}})^2 = -{{{31/64}}}
s + {{{1/8}}} = +/-{{{sqrt(-31/64)}}}
s = -{{{1/8}}} +/-{{{sqrt(-31/64)}}}
Extract square root of -1 and square root of (1/64)
s = -{{{1/8}}} +/-{{{1/8}}}i{{{sqrt(31)}}}
Can also be written
s = {{{(-1 + sqrt(31)i)/8}}}
and
s = {{{(-1 - sqrt(31)i)/8}}}
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Was this sufficient steps?