# 4r^2-28r=-49 , solve with the quadratic formula

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### 2 Answers

`4r^2-28r=-49`

`4r^2-28r+49=0`

We wish to solve the above problem.

Let us factorize the problem

`4x^2-14x-14x+49=0`

`2x(2x-7)-7(2x-7)=0`

`(2x-7)(2x-7)=0`

`So`

`2x-7=0`

`2x=7`

`x=7/2`

OR

solving with quadratic formula i.e. if `ax^2+bx+c=0` is quadratic equation then

`x=(-b+-sqrt(b^2-4ac))/(2a)` ,

therefore

`4r^2-28r+49=0`

Thus

a=4,b=-28 and c=49

`r=(-(-28)+-sqrt((-28)^2-4xx4xx49))/(2xx4)`

`r=(28+-0)/(8)`

`r=7/2`

Ans.

4r^2-28r=-49

(4r^2-14r)-(14r+49)=0

2r(2r-7)-7(2r-7)=0

(2r-7)(2r-7)=0

2r-7=0

2r=7

r=7/2