Question

easiest way to solve this 4got how to solve
use quadratic formula to solve this question round solutions to nearest hundereth if neccesary


1) -3x^2-4x+10=0

2) x^2+6x-100=0

Answer 1

\[x^2+6x=100\]
\[(x+3)^2=100+9=109\]
\[x+3=\pm\sqrt{109}\]
\[x=-3\pm\sqrt{109}\]

Answer 2

THANK YOU Do you know the first one

Answer 3

\[ -3x^2-4x+10=0\]
\[3x^2+4x-10=0\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] witih \(a=3,b=4,c=-10\)