Question

How to solve 7r^2-41r-6=0 basics step by step

Answer 1

first,press the eqn
then,
choose number 2,
then,
a=7,b=-41,c=-6
(r-6) (r=-0.143)
r=6,r=0.143

Answer 2

We are trying to solve this equation

`7r^(2) -41r-6=0`

We need to split up -41r in order to factor by grouping. We need to spilt -41r into 2, where one can be factored with
`7r^(2)` and other part can be factored with -6.

We can find out how to split it by multipling 7 and -6 to get -42. We need to split -41 where it can be multiplied to get -42. It is -42 and 1.

`7r^(2) +r-42r-6=0`

We can then factor by grouping them

With
`7r^(2)` and r, they can factor out a r.

With -42r and -6, they can factor out a -6.

So it will be
`r(7r+1)-6(7r+1)=0`

Notice that inside the parenthesizes, both are 7r+1, we want it to be the same.

We can arrange it into (r-6)(7r+1)=0

We need to find what values of r would make the equation equal to 0.

Multiplying anything by 0 will equal 0, so there will be 2 solutions.

For (r-6), if r=6 then the equation will = 0

For (7r+1), if r = -1/7 then the equation will = 0.

So the solution will be r= -1/7, 6