Yes. You can, too.

For quadratic

a*x^2 + b*x + c = 0

The solution is

x = (-b ±√(b^2 - 4*a*c))/(2*a) (this is the quadratic formula)

You have a=2, b=-9, c=10. Making the substitutions, we get

x = (-(-9) ±√((-9)^2 - 4*2*10))/(2*2)

= (9 ±√(81-80))/4

= (9±1)/4

= {10/4, 8/4}

For quadratic

a*x^2 + b*x + c = 0

The solution is

x = (-b ±√(b^2 - 4*a*c))/(2*a) (this is the quadratic formula)

You have a=2, b=-9, c=10. Making the substitutions, we get

x = (-(-9) ±√((-9)^2 - 4*2*10))/(2*2)

= (9 ±√(81-80))/4

= (9±1)/4

= {10/4, 8/4}

**x = 2 1/2 or 2**