Question 74972
{{{2t^2+7t=4}}} set = to 0
{{{2t^2+7t-4=4-4}}}
{{{2t^2+7t-4=0}}} Factor however you know how, I'm using grouping or ac method.
2t^2+__t ___t-4=0
Fill in the blanks with two numbers that multiply to give you the product of the first and last coefficients (2*-4=-8), but add to give you the middle coefficient (7).  8*-1=-8 and 8+(-1)=7
{{{2t^2+8t-1t-4=0}}}  Factor the GCF out of the first two terms and the last two terms:
{{{2t(t+4)-1(t+4)=0}}} factor out (t+4)
{{{(t+4)(2t-1)=0}}}  set each parentheses=0 and solve for t.
t+4=0 or 2t-1=0
t+4-4=0-4 or 2t-1+1=0+1
t=-4 or 2t=1
t=-4 or 2t/2=1/2
t=-4 or t=1/2
Check by letting t=-4  and t=1/2 and see if both sides equal.
for t=-4
{{{2(-4)^2+7(-4)=4}}}
{{{2(16)-28=4}}}
{{{32-28=4}}}
{{{4=4}}}  This one works.
for t=1/2
{{{2(1/2)^2+7(1/2)=4}}}
{{{2(1/4)+7(1/2)=4}}}
{{{2/4+7/2=4}}}
{{{1/2+7/2=4}}}
{{{(1+7)/2=4}}}
{{{8/2=4}}}
{{{4=4}}}  This one works too.
Happy Calculating!!!!