Question 151055
{{{2x^2-4x=3}}}
{{{2x^2-4x-3=0}}}
Now you can use the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-4) +- sqrt((-4)^2-4*2*(-3) ))/(2*2) }}} 
{{{x = (4 +- sqrt(16+24))/(4) }}} 
{{{x = (4 +- sqrt(40))/(4) }}}
{{{x = (4 +- 2sqrt(10))/(4) }}} 
{{{x = 1 +- sqrt(10)/2 }}} 
Is that what you meant by sqrt(10)?
Here are the approximate solutions,
{{{x[1] = 1 + sqrt(10)/2 }}}
{{{x[1] = 1 + 3.162/2 }}}
{{{x[1] = 1 + 1.581 }}}
{{{x[1] = 2.581 }}}
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{{{x[2] = 1 - sqrt(10)/2 }}}
{{{x[2] = 1 - 1.581 }}}
{{{x[2] = -0.581 }}}
{{{drawing( 300, 300, -5, 5, -5, 5,grid( 1 ),circle( 2.581, 0, 0.2 ),circle( -.581, 0, 0.2 ), graph( 300, 300, -5, 5, -5, 5, 2x^2-4x-3)) }}}