Answer
$\left\{\dfrac{-3-\sqrt{265}}{8},\dfrac{-3+\sqrt{265}}{8}\right\}$
Work Step by Step
We are given the equation:
$\dfrac{1}{2}x^2+\dfrac{3}{8}x=2$
Multiply all terms by 8 to clear denominators, then write the equation in standard form:
$8\cdot\dfrac{1}{2}x^2+8\cdot\dfrac{3}{8}x=8\cdot2$
$4x^2+3x=16$
$4x^2+3x-16=0$
Solve the equation using the quadratic formula:
$x=\dfrac{-3\pm\sqrt{3^2-4(4)(-16)}}{2(4)}=\dfrac{-3\pm\sqrt{265}}{8}$
$x_1=\dfrac{-3-\sqrt{265}}{8}$
$x_2=\dfrac{-3+\sqrt{265}}{8}$
The solution set of the equation is:
$\left\{\dfrac{-3-\sqrt{265}}{8},\dfrac{-3+\sqrt{265}}{8}\right\}$
