Solve 2x2 + 3x + 6 = 0. Round solutions to the nearest hundredth.

x ≈ -2.64 and x ≈ 1.14
x ≈ -1.14 and x ≈ 2.64
No real solutions
x ≈ -1.11 and x ≈ -4.89

Question

Solve 2x2 + 3x + 6 = 0. Round solutions to the nearest hundredth.

x ≈ -2.64 and x ≈ 1.14
x ≈ -1.14 and x ≈ 2.64
No real solutions
x ≈ -1.11 and x ≈ -4.89

anonymous · Answer

Did you mean?
$$2x^2+3x+6=0$$
If so, use this: 
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ Better? :-)

anonymous · Answer

$$ax^2 + bx + c$$
where a, b, & , c are the constants

ashking1 · Answer

i really dont understand i need it quick for a test

anonymous · Answer

Shortcut, use the discriminant part of the quadratic equation:

b² − 4ac > 0 ; 2 real roots
b² − 4ac = 0 ; 1 double real root
b² − 4ac < 0 ; 2 imaginary/non-real roots

The answer should be obvious now I believe, yes?

anonymous · Answer

a = 2, b = 3 , c = 6

ashking1 · Answer

ok i got it thks

anonymous · Answer

Aye then good work! :D  Hope that makes more sense now