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

Answer 1

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

Answer 2

\[ax^2 + bx + c\]
where a, b, & , c are the constants

Answer 3

i really dont understand i need it quick for a test

Answer 4

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?

Answer 5

a = 2, b = 3 , c = 6

Answer 6

ok i got it thks

Answer 7

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