Question 52064
{{{y^2-6y=16}}}
{{{y^2 -6y + 9 = 16 + 9}}}
{{{(y - 3)^2 = 25}}}
What I did was take 1/2 of the coefficient of y, square it, and add it to 
both sides
{{{y - 3 = 5}}}
{{{y = 8}}}
check
{{{y^2-6y=16}}}
{{{8^2 -6*8 = 16}}}
{{{64 - 48 = 16}}}
{{{16 = 16}}}
OK
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{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
If  b^2-4*a*c turns out to be positive, then the solutions are real
If it's negative, the solutions are imaginary
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3p^2-3p-4=0
{{{p = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{p = (-(-3) +- sqrt( (-3)^2-4*3*(-4) ))/(2*3) }}}
{{{p = (3 +- sqrt(57 ))/6 }}} answer
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2x^2-5x-12=0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (5 +- sqrt( 25-96 ))/4 }}}
{{{x = (5 +- sqrt( 121 ))/4 }}}
{{{x = (5 + 11)/ 4}}}
{{{x = (5 - 11)/ 4}}}
So the solutions are x = 4, x = - 3/2