Question

]2x^2-5x+2=0 como resolver?

Answer 1

$2x^2-5x+2=0\\ \\ \Delta=(-5)^2-4.2.2=25-16=9\\ \\ x=\frac{5\pm\sqrt9}{2.2}=\frac{5\pm3}{4}\\ \\ x_1=\frac{5-3}{2}=1\\ \\ x-2=\frac{5+3}{2}=4$