Question

PLEASE PLEASE HELP

Part A::(subtract the polynomials)
(SHOW ALL WORK)!!

(3-6n^5-8n^4) - (-6n^4-3n-8n^5)

(Part B:: solve the quadratic)

SHOW ALL WORK
4x^2-2x-5=0

Thanks

Answer 1

$(3-6n^5-8n^4) - (-6n^4-3n-8n^5)\\\\=3-6n^5-8n^4+6n^4+3n+8n^5\\\\=(8n^5-6n^5)+(6n^4-8n^4)+3n+3\\\\=\boxed{2n^5-2n^4+3n+3}$


$4x^2-2x-5=0\\a=4;\ b=-2;\ c=-5\\\\\Delta=b^2-4ac\\\Delta=(-2)^2-4\cdot4\cdot(-5)=4+80=84\\\sqrt\Delta=\sqrt{84}=\sqrt{4\cdot21}=2\sqrt{21}\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-(-2)-2\sqrt{21}}{2\cdot4}=\dfrac{2-2\sqrt{21}}{8}=\dfrac{1-\sqrt{21}}{4}\\\\x_2=\dfrac{-(-2)+2\sqrt{21}}{2\cdot4}=\dfrac{2+2\sqrt{21}}{8}=\dfrac{1+\sqrt{21}}{4}$