Question

Find the real or imaginary solution of each equation by factoring.

2x^2+3=4x

Answer 1

$2x^2+3=4x \\\\2x^2-4x+3=0\\\\\Delta=(-4)^2-4\cdot2\cdot3=16-24=8\\\\\sqrt{\Delta}=\sqrt8=2\sqrt2\\\\x_1=\dfrac{-(-4)-2\sqrt2}{2\cdot2}=\dfrac{2-\sqrt2}{2}\\\\x_2=\dfrac{-(-4)+2\sqrt2}{2\cdot2}=\dfrac{2+\sqrt2}{2}$