I' not really sure how to solve this one I would greatly appreciate the help
[9.05] 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

I' not really sure how to solve this one I would greatly appreciate the help
[9.05] 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

we can use the formula for qudratic equtions of the type ax^2+bx+c=0

anonymous · Answer

$$x=(-b \pm(\sqrt{b^2-4(a*c)})/2a$$

anonymous · Answer

this has no real solutions

anonymous · Answer

thank you but can you explain why please im not familiar with the formula

anonymous · Answer

ok..you saw that sqrt(b^2-4(a*c) part?

anonymous · Answer

yeah

anonymous · Answer

we have b^2=3^2=9. 4(a*c) will be 4*2*6=48

anonymous · Answer

then the element inside sqrt.() will be negative...there is no root for a neg.number

anonymous · Answer

thank you very much i really appreciate the help

anonymous · Answer

you are welcome:D