Question 75083
The quadratic equation is {{{2x^2-5x-3=0}}}.
Comparing with {{{ax^2+bx+c=0}}}, a = 2, b = -5, c = -3.
The quadratic formula is
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
So here {{{x = (-(-5) +- sqrt( (-5)^2-4*2*(-3) ))/(2*2) }}}
{{{x = (5 +- sqrt( 25+4*2*3 ))/4 }}}
{{{x = (5 +- sqrt( 25+24 ))/4 }}}
{{{x = (5 +- sqrt( 49 ))/4 }}}
{{{x = (5 +- 7)/4 }}}
Either {{{x = (5 + 7)/4 }}} or {{{x = (5 - 7)/4 }}}
Either {{{x = 12/4 }}} or {{{x = -2/4 }}}
Either {{{x = 3 }}} or {{{x = -1/2 }}}


Your answer is wrong!


Remark: Check the answer by factoring [Hint: (2x+1)(x-3) = 0].