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

Mar 13, 2016

Factor by grouping to find zeros:

$x = \frac{4}{7}$ and $x = 7$

#### Explanation:

For subtract $53 x$ from both sides to get:

$7 {x}^{2} - 53 x + 28 = 0$

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

$0 = 7 {x}^{2} - 53 x + 28$

$= 7 {x}^{2} - 49 x - 4 x + 28$

$= \left(7 {x}^{2} - 49 x\right) - \left(4 x - 28\right)$

$= 7 x \left(x - 7\right) - 4 \left(x - 7\right)$

$= \left(7 x - 4\right) \left(x - 7\right)$

Hence zeros $x = \frac{4}{7}$ and $x = 7$

Mar 13, 2016

(4/7) and 7.

#### Explanation:

I use the new Transforming Method (Google, Yahoo Search).
$y = 7 {x}^{2} - 53 x + 28 = 0$
Transformed equation: $y ' = {x}^{2} - 53 x + 196$
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: $\left(\frac{4}{7}\right)$ and
$\left(\frac{49}{7} = 7\right)$

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