$x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\y=x^2+3x+5\\\\a=1;\ b=3;\ c=5\\\\b^2-4ac=3^2-4(1)(5)=9-20=-11\\\\x_1=\dfrac{-3-\sqrt{-11}}{2\cdot1}=\dfrac{-3-\sqrt{-11}}{2}\\\\x_2=\dfrac{-3+\sqrt{-11}}{2}$

Answer: A and B.

3 0

3 years ago

Read 2 more answers

Hello hope you are well can I have some help on this factorise question please 4f2 + f

lapo4ka [179]

Answer:

f(4f + 1)

Step-by-step explanation:

Given

4f² + f ← factor out f from each term

= f(4f + 1)

6 0

3 years ago

Should any factors be accounted for when explaining how to solve an equation?

Veronika [31]

Yes, it should be taken into account

3 0

3 years ago