Question

HELP ME!!! ITS DUE TOMMOROW!! Find the largest value of $r$ such that $6r^2+7r+8=6$.

Answer 1

$6r^{2}+7r+8=6\\\\6r^{2}+7r+2=0\\\\(3x+2)(2x+1)=0\\\\ 3x+2=0\\\\3x=-2\\\\x=\frac{-2}{3}\\\\2x+1=0\\\\2x=-1\\\\x=\frac{-1}{2}$

Since r = -2/3 and r = -1/2, the answer is -1/2 because it is larger than -2/3.

Answer 2

First you need to see that this is a quadratic. I need.to put all of the values on one side of equation to see what I got.
6r^2 + 7r + 8 = 6
6r^2 + 7r + 2 = 0

Now this one is difficult to factor so i will use quadratic equation:
[-b (+-) sqrt (b^2 - 4ac)] / (2a)

we know that a b and c are in a quadratic at these positions.
ax^2 + bx + c

so
[-7 (+-) sqrt (7^2 -(4)(6)(2)] / (2) (6)
[-7 (+-) sqrt (49 - 48)] / 12
[-7 (+-) 1] /12

split into the + and - for 2 answers
(-7 + 1) / 12
-6/12
-1/2

And
(-7 -1) /12
-8 / 12
-2/3

those are.the 2 answers

But but it says largest so -1/2