Question

How to solve 2x^2 +7x +3 =0By undoing

Answer 1

From the problem :

`\begin{gathered} 2x^2+7x+3=0 \\ \end{gathered}`

We need to have a perfect square trinomial, rewriting the equation :

`\begin{gathered} (2x^2+7x)+3=0 \\ (2x^2+7x+c)+3=0+c \end{gathered}`

We will add a term c to both sides of the equation.

Factor out 2 so that the coefficient of x^2 will be 1

`2(x^2+(7)/(2)x+(c)/(2))+3=0+c`

The term c/2 will be equal to the value of half the middle term and squared.

So this will be :