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

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

`x=(-57pm sqrt(57^2-4*14*28))/(2*14)=(-57pm sqrt(3249-1568))/28`

`=(-57pm sqrt(1681))/28=(-57 pm 41)/28`

Now `(-57+41)/28=-16/28=-4/7` and `(-57-41)/28=-98/28=-7/2`.

(2) Factoring can also be done as follows:

`14x^2+57x+28=0\rightarrow (2x+7)(7x+4)=0`, which leads to the same answers.

(3) Completing the square can be done as follows. Note that `3249/784=(57/28)^2`.

`14x^2+57x+28=0\rightarrow 14(x^2+57/14 x+3249/784)+28=14*3249/784`

`\rightarrow 14(x+57/28)^2=1681/56\rightarrow (x+57/28)^2=1681/784`

`\rightarrow x+57/28=pm sqrt(1681/784)= pm 41/28`

`\rightarrow x=(-57pm 41)/28`

As above, `(-57+41)/28=-16/28=-4/7` and `(-57-41)/28=-98/28=-7/2`.

I use the new Transforming Method (Socratic Search)
Transformed equation: `y' = x^2 + 57x + 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: `x1 = -8/14 = -4/7` and `x2 = -49/14 = -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.