How do you factor and solve\[2{{x}^{2}}+15x+7=0\]?

Hint: First of all we should know about the formula for quadratic equation \[\text{a}{{\text{x}}^{2}}+bx+c\] and roots for that equation i.e.\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Then we have to compare the values of a, b, c, with our equation. To get a solution for the given problem we have to find roots for the given equation. For that we have to substitute the values of a, b, c in the roots equation.


Complete step by step answer:
To find the factor and solution for \[2{{x}^{2}}+15x+7=0\] .
Let us consider the given equation as equation (1)
\[2{{x}^{2}}+15x+7=0.........\left( 1 \right)\]
By observing the power of equation (1) we can clearly say that the given equation is a quadratic equation.
As we know that
\[\begin{align}
  & For\text{ a}{{\text{x}}^{2}}+bx+c=0; \\
 & \text{The roots of the equation are }\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ } \\
\end{align}\]
Let us consider
\[\text{a}{{\text{x}}^{2}}+bx+c=0........\left( f1 \right)\]
\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}...........(f2)\]
For getting a, b, c values let us compare equation (1) with formula (f1).
By comparing equation (1) with formula (f2) we will get
\[a=2\text{ b=15 c=7}\]
Now we have to submit above a, b, c values in formula (f2) , we get

\[\dfrac{-\left( 15 \right)\pm \sqrt{{{\left( 15 \right)}^{2}}-4\left( 2 \right)\left( 7 \right)}}{2\left( 2 \right)}\]
By simplifying the above equation, we get
\[\Rightarrow \dfrac{-15\pm \sqrt{225-56}}{4}\]
Subtracting 225-56 and then, we get
\[\Rightarrow \dfrac{-15\pm \sqrt{169}}{2}\]
Rewriting 169 as 13.13, we get
\[\Rightarrow \dfrac{-15\pm \sqrt{13.13}}{2}\]
Getting out 13 from the square root, we have
\[\Rightarrow \dfrac{-15\pm 13}{2}\]
Therefore factors or roots of the equation (1) is
\[\Rightarrow \dfrac{-15\pm 13}{2}\]
Let us consider \[\dfrac{-15\pm 13}{2}\] as equation (2).
\[\Rightarrow {{R}_{12}}=\dfrac{-15\pm 13}{2}.............\left( 2 \right)\]
Taking + case in equation (2), we get
\[\begin{align}
  & \Rightarrow \dfrac{-15+13}{2} \\
 & \Rightarrow \dfrac{-2}{2} \\
 & \Rightarrow -1 \\
\end{align}\]
Let us consider \[{{R}_{1}}=-1\]as equation (3)
\[{{R}_{1}}=-1..........\left( 3 \right)\]
Taking – case in equation (3), we get
\[\begin{align}
  & \Rightarrow \dfrac{-15-13}{2} \\
 & \Rightarrow \dfrac{-28}{2} \\
 & \Rightarrow -14 \\
\end{align}\]
Let us consider \[{{R}_{2}}=-14\] as equation (4), we get
\[{{R}_{2}}=-14.......\left( 4 \right)\]
So, therefore equation (3) and (4) are the factors of equation (1).

Note:
 Students should have good knowledge of quadratic equation concepts. They have to solve multiple examples of different models to get a good grip on this concept. Sometimes the examiner may give both roots and ask to write quadratic equations from the roots. So, students should be aware of all types of problems in this concept.