# How do you solve 14 x ^2+ 57x + 28 = 0?

Oct 9, 2015

$x = - \frac{7}{2}$ and $x = - \frac{4}{7}$ are the two solutions. They can be found by the quadratic formula, by factoring, or by completing the square.

#### Explanation:

(1) For $14 {x}^{2} + 57 x + 28 = 0$, the quadratic formula gives:

$x = \frac{- 57 \pm \sqrt{{57}^{2} - 4 \cdot 14 \cdot 28}}{2 \cdot 14} = \frac{- 57 \pm \sqrt{3249 - 1568}}{28}$

$= \frac{- 57 \pm \sqrt{1681}}{28} = \frac{- 57 \pm 41}{28}$

Now $\frac{- 57 + 41}{28} = - \frac{16}{28} = - \frac{4}{7}$ and $\frac{- 57 - 41}{28} = - \frac{98}{28} = - \frac{7}{2}$.

(2) Factoring can also be done as follows:

$14 {x}^{2} + 57 x + 28 = 0 \setminus \rightarrow \left(2 x + 7\right) \left(7 x + 4\right) = 0$, which leads to the same answers.

(3) Completing the square can be done as follows. Note that $\frac{3249}{784} = {\left(\frac{57}{28}\right)}^{2}$.

$14 {x}^{2} + 57 x + 28 = 0 \setminus \rightarrow 14 \left({x}^{2} + \frac{57}{14} x + \frac{3249}{784}\right) + 28 = 14 \cdot \frac{3249}{784}$

$\setminus \rightarrow 14 {\left(x + \frac{57}{28}\right)}^{2} = \frac{1681}{56} \setminus \rightarrow {\left(x + \frac{57}{28}\right)}^{2} = \frac{1681}{784}$

$\setminus \rightarrow x + \frac{57}{28} = \pm \sqrt{\frac{1681}{784}} = \pm \frac{41}{28}$

$\setminus \rightarrow x = \frac{- 57 \pm 41}{28}$

As above, $\frac{- 57 + 41}{28} = - \frac{16}{28} = - \frac{4}{7}$ and $\frac{- 57 - 41}{28} = - \frac{98}{28} = - \frac{7}{2}$.

Oct 9, 2015

Solve: $y = 14 {x}^{2} + 57 x + 28 = 0$ (1)

Ans $x = - \frac{4}{7}$ and $x = - \frac{7}{2}$

#### Explanation:

I use the new Transforming Method (Socratic Search)
Transformed equation: $y ' = {x}^{2} + 57 x + 392 = 0$ (2).
Both roots are negative. Factor pairs of (392) --> (-2, -196)(-4, -98)(-8, -49). This sum is -57 = -b. Then the 2 real roots of (2) are: -8 and -49.
Therefor, the 2 real roots of original equation (1) are: $x 1 = - \frac{8}{14} = - \frac{4}{7}$ and $x 2 = - \frac{49}{14} = - \frac{7}{2.}$

NOTE . Solving by this Transforming Method is simpler and faster because it avoids the lengthy factoring by grouping, or the boring computation with the formula. In addition, it avoids solving the 2 binomials.