How do you solve \[2{{x}^{2}}+9x+7=0\] ?

Hint: We can solve this question either by completing the square or by using a quadratic formula. To solve this question by the method of completing the square we will follow a few steps. First of all, write the given equation in decreasing order of the degree of \[x\]. then separate the constant or the number to the right side of the equation. Now, divide all the terms with the coefficient of \[{{x}^{2}}\] to make the coefficient of \[{{x}^{2}}\] one. After that we have to divide the coefficient of \[x\] by two and then square it, add this value to both sides of the equation. Now, take the square root on both sides. After that simplify to get the value of \[x\] .

Complete step by step answer:
In this question the given equation is \[2{{x}^{2}}+9x+7=0\]
Since, the coefficient of \[{{x}^{2}}\] is two, so have to divide both the sides by two in order to make the coefficient one. Now moving the constant term to the other side and dividing the coefficient of by two and adding value on both sides.
\[\begin{align}
  & 2{{x}^{2}}+7x+9=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+\dfrac{9}{2}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x=-\dfrac{9}{2} \\
 & \\
\end{align}\]
Here the value of \[\dfrac{b}{2}=\dfrac{7}{4}\]
\[\begin{align}
  & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+{{\left( \dfrac{7}{4} \right)}^{2}}={{\left( \dfrac{7}{4} \right)}^{2}}-\dfrac{9}{2} \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+{{\left( \dfrac{7}{4} \right)}^{2}}={{\left( \dfrac{49}{16} \right)}^{{}}}-\dfrac{9}{2} \\
 & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}=\left( \dfrac{49}{16} \right)-\dfrac{9}{2} \\
 & \Rightarrow {{\left( x+\dfrac{7}{4} \right)}^{2}}=-\left( \dfrac{23}{16} \right) \\
 & \\
\end{align}\]
Now, taking the square root on both sides.
\[\Rightarrow \left( x+\dfrac{7}{4} \right)=\sqrt{-\left( \dfrac{23}{16} \right)}\]
Here we can see that the roots of this equation are imaginary because the value inside the root is negative.
Therefore using iota we will simplify
\[\begin{align}
  & \left( x+\dfrac{7}{4} \right)=\sqrt{-\left( \dfrac{23}{16} \right)} \\
 & \Rightarrow x=\sqrt{-\left( \dfrac{23}{16} \right)}-\dfrac{7}{4} \\
 & \Rightarrow x=i\sqrt{\left( \dfrac{23}{16} \right)}-\dfrac{7}{4} \\
 & \Rightarrow x=-1.75\pm 1.19895788i \\
\end{align}\]

Hence after solving we got \[x=-1.75\pm 1.19895788i\] .

Note:
We can solve this question using a quadratic formula which is \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . By simply substituting the values we can get the roots. While applying square root on both sides don’t forget to add both the signs as value from the root is always positive regardless of the number inside the square root. Also perform the calculations carefully in order to avoid silly mistakes.