How do you factor \[2{{x}^{2}}+3x=5\] ?

Hint: In the given question, we have been asked to factorize the given quadratic equation. In order to solve the question, we first need to write the equation in the standard form i.e. \[a{{x}^{2}}+bx+c=0\], then splitting the middle term into two factors and solve the further equation to get the factors of the given quadratic equation.

Formula used:
The standard form of the quadratic equation is:
\[a{{x}^{2}}+bx+c=0\]

Complete step by step answer:
We have the given quadratic equation:
\[2{{x}^{2}}+3x=5\]
Move terms to the left side, we get
\[2{{x}^{2}}+3x-5\]
Equate the equation with zero, we get
\[2{{x}^{2}}+3x-5\]
Split the middle term by using sum-product pattern:
\[2{{x}^{2}}+5x-2x-5\]
Taking common factors from the two pairs, we get
\[\left( 2{{x}^{2}}+5x \right)+\left( -2x+5 \right)\]
\[\Rightarrow x\left( 2x+5 \right)-1\left( 2x+5 \right)\]
Rewrite in the factored form, we get
\[\therefore\left( x-1 \right)\left( 2x+5 \right)\]

Therefore, \[\left( x-1 \right)\ and\ \left( 2x+5 \right)\] are the common factors of the given equation.

Additional information:
If question asked to solve for the value of \[x\]:
There are the following steps need to be followed further after factorization:
Equate each common factor pair with 0, we get
\[x-1=0\] and \[2x+5=0\]
Solve for the value of \[x\], we get
\[x=1,\ -\dfrac{5}{2}\]
Therefore, the value of \[x\] are \[1\ and\,-\dfrac{5}{2}\].\[\]

Note:In this question, it is important to note that the factorization method by splitting the middle term works for all quadratic equations. The standard form of the quadratic equation is \[a{{x}^{2}}+bx+c=0\]. If the question is to solve the equation, then we need to equate the common factor is equal to 0 and solve for the value of the given variable.Finding a factor is to find an expression or a number or a term which we can multiply by.