Answer
$x_{1}=\dfrac{2 + \sqrt{10}}{2}$ and $x_{2}=\dfrac{2- \sqrt{10}}{2}$
Work Step by Step
Given $2x^2-4x=3 \longrightarrow 2x^2-4x-3=0$
$a = 2, \ b = -4, \ c = -3$
Using the quadratic formula $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-4)\pm \sqrt{(-4)^2-4\times 2\times (-3)}}{2\times2} = \dfrac{4\pm \sqrt{16+24}}{4} = \dfrac{4\pm \sqrt{40}}{4} = \dfrac{4\pm 2\sqrt{10}}{4} = \dfrac{2\pm \sqrt{10}}{2}$
Therefore, the solutions are $x_{1}=\dfrac{2 + \sqrt{10}}{2}$ and $x_{2}=\dfrac{2- \sqrt{10}}{2}$
