How do you solve $2{{x}^{2}}+6x=10$ ?

Hint: We can solve for the solution of the given equation by applying the complete square method. We will add numbers on both the sides after rearranging some terms to have a complete square in the form of ${{\left( x+a \right)}^{2}}$ . Now, further simplifying the equation we will be able to find the solution of the equation.

Complete step-by-step solution:
The given equation is
$2{{x}^{2}}+6x=10$
We now divide both sides of the above equation by $2$ as
$\Rightarrow {{x}^{2}}+3x=5$
We rewrite the equation as
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{3}{2}x=5....\text{expression}1$
To have a complete square in the left hand side, let’s take the square ${{\left( x+a \right)}^{2}}$ for comparison.
We know that, ${{\left( x+a \right)}^{2}}={{x}^{2}}+2\cdot a\cdot x+{{a}^{2}}....\text{expression2}$
As, the first two terms in the expression $\left( {{x}^{2}}+2\cdot a\cdot x+{{a}^{2}} \right)$ are ${{x}^{2}}$ and $2x$ , we compare the left-hand side of $\text{expression}1$ with the right-hand side of $\text{expression2}$ and get the value of $a$ as $a=\dfrac{3}{2}$ .
Hence, to get the square term ${{\left( x+\dfrac{3}{2} \right)}^{2}}$ we add $\dfrac{9}{4}$ to the both sides of $\text{expression}1$
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{3}{2}x+\dfrac{9}{4}=5+\dfrac{9}{4}$
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{3}{2}x+\dfrac{9}{4}=\dfrac{29}{4}$
The above equation can be also written as
$\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{29}{4}$
Now, taking square root on both the sides of the equation, we get
$\Rightarrow x+\dfrac{3}{2}=\pm \dfrac{\sqrt{29}}{2}$
Further simplifying the above equation, we get
$\Rightarrow x=\dfrac{\sqrt{29}}{2}-\dfrac{3}{2}$
$\Rightarrow x=\dfrac{\sqrt{29}-3}{2}$
And $\Rightarrow x=-\dfrac{\sqrt{29}}{2}-\dfrac{3}{2}$
$\Rightarrow x=-\dfrac{\sqrt{29}+3}{2}$
Therefore, the solution of the equation $2{{x}^{2}}+6x=10$ are $x=-\dfrac{\sqrt{29}+3}{2}$ and $x=\dfrac{\sqrt{29}-3}{2}$.

Note: We have to keep in mind while simplifying the last part of the solution we have to take both the positive and negative signs of $\dfrac{\sqrt{29}}{2}$ , otherwise we will not be getting both the solutions. The given problem can also be solved by other methods, such as by factoring the left-hand side of the equation and equating both the factors individually to zero. Also, it can be done by using directly Sridhar Acharya formula to find the solution, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .