How do you solve $6{y^2} + 11y - 21?$

Hint: In this question we have to find the roots from the above quadratic equation by quadratic formula. For that we are going to simplify the equation. Next, we extract the square root of both sides and then simplify to arrive at our final answer. And also we are going to add and subtraction in complete step by step solution.

Complete step by step solution:
The quadratic formula is ${y_{1,2}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
To find the roots from the given quadratic equation:
Given,
$ \Rightarrow 6{y^2} + 11y - 21$
First, we take $a = 6$,$b = 11$, and $c = - 21$ from the above equation.
Next, we substitute the values in the quadratic formula and we get
\[ \Rightarrow {y_{1,2}} = \dfrac{{ - 11 \pm \sqrt {{{( - 11)}^2} - 4 \times 6 \times ( - 21)} }}{{2 \times 6}}\]
Next, simplify the above term and we get
\[ \Rightarrow {y_{1,2}} = \dfrac{{ - 11 \pm \sqrt {121 + 504} }}{{12}}\]
Adding the values inside of the square values and we get
\[ \Rightarrow {y_{1,2}} = \dfrac{{ - 11 \pm \sqrt {625} }}{{12}}\]
Here, $\sqrt {625} = 25$
Put the value in the above term
\[ \Rightarrow {y_{1,2}} = \dfrac{{ - 11 \pm 25}}{{12}}\]
Next, we split the above term and we get
\[ \Rightarrow {y_1} = \dfrac{{ - 11 + 25}}{{12}}\]
\[ \Rightarrow {y_2} = \dfrac{{ - 11 - 25}}{{12}}\]
Finally, we adding and subtract the numerator values and we get the required factor:
 \[ \Rightarrow {y_1} = \dfrac{{14}}{{12}}\]
\[ \Rightarrow {y_2} = \dfrac{{ - 36}}{{12}}\]
Next, divide the above term
\[ \Rightarrow {y_1} = \dfrac{7}{6}\]
\[ \Rightarrow {y_2} = - 3\]
This is the required answer.

Note: Factorization of a quadratic expression is the opposite of expansion, and is the process of putting the brackets back into the expression rather than taking them out. To factorize a quadratic expression of the form \[a{x^2} + bx + c\;\] you must find two numbers that add together to give the first coefficient of x and multiply to give the second coefficient of x.
There is another little hard way to find the answer.
I use the new AC Method to factor trinomials (Socratic Search)
\[ \Rightarrow f\left( y \right){\text{ }} = {\text{ }}6\left( {y{\text{ }} + {\text{ }}p} \right)\left( {y{\text{ }} + {\text{ }}q} \right)\]
Converted trinomial:
\[ \Rightarrow y' = {x^2} + 11y - 126\]
\[ \Rightarrow y' = \;\left( {y{\text{ }} + {\text{ }}p'} \right)\left( {y{\text{ }} + {\text{ }}q'} \right).\]
Here,\[p'\] and \[q'\] have opposite signs.
Factor pairs of \[\left( { - 126} \right) = \left( { - 2,{\text{ }}63} \right)\left( { - 3,{\text{ }}42} \right)\left( { - 6,{\text{ }}21} \right)\left( { - 7,{\text{ }}18} \right).\]
This sum is \[11{\text{ = }}b\]. Then \[p'{\text{ = }} - 7\] and \[q' = {\text{ }}18\].
Therefore,\[\;p = p',a = - 76\] and \[q = q',a = 186 = 3\]
Factored form:
$ \Rightarrow y = 6(x - 76)(x + 3)$
\[ \Rightarrow (6x - 7)(x + 3)\]
$ \Rightarrow x = \dfrac{7}{6}, - 3$
This is the required answer.
It is used in bond trading and mortgage calculations. The polynomial is of high order, for example, with an interest term with exponent 360 for a 30-year mortgage. This is not a formula that can be factored. Instead, if the interest needs to be calculated, it is solved by computer or calculator.