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use the quadratic formula to solve : 2x^2-5x=1
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\[5\pm \sqrt{17}/4\]
\[-b \pm \sqrt{b^2-4ac}/2a is your general form of \] is your general form of quadratic eq
is that... \[5+\sqrt{17}/4 \] as in 5+or- (sqrt) 17 over 4?
yeah
is the answer satisfactory ?
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2x^2 -5x =1 2x^2 -5x -1 =0 x_1,2 = (5+/- sqrt(25+8)/4 = (5+/- sqrt33)/4
yes it is
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