# How do you solve #7x^2 +28=53x#?

##### 2 Answers

Mar 13, 2016

Factor by grouping to find zeros:

#x = 4/7# and#x=7#

#### Explanation:

For subtract

#7x^2-53x+28 = 0#

We can split the middle term and factor by grouping as follows:

#0 =7x^2-53x+28#

#=7x^2-49x-4x+28#

#=(7x^2-49x)-(4x-28)#

#=7x(x-7)-4(x-7)#

#=(7x-4)(x-7)#

Hence zeros

Mar 13, 2016

(4/7) and 7.

#### Explanation:

I use the new Transforming Method (Google, Yahoo Search).

Transformed equation:

Factor pairs of (ac = 196) --> (2, 98)(4, 49). This sum is 53 = -b. Then the 2 real roots of y' are: 4 and 49.

Back to original equation y, the 2 real roots are:

**NOTE** . There is no need to factor by grouping and solving the 2 binomials.