Question

Find the real solutions of the equation$2y^2+7y+3=0$by factoring.

Short answer

The real solution of the equation is $y=-\frac{1}{2}\text{and}y=-3$.

Step-by-step solution

Step 1. Factorized the given quadratic equation.

$$\begin{gathered} 2y^2+7y+3=0....\left(\text{Given}\right) \\ 2y^2+\left(y+6y\right)+3=0 \\ y\left(2y+1\right)+3\left(2y+1\right)=0....\left(\text{Factorized}\right) \\ \left(2y+1\right)\left(y+3\right)=0....\left(1\right) \end{gathered}$$

Step 2. Further simplifying (1).

$$\begin{gathered} \left(2y+1\right)\left(y+3\right)=0....\left(\text{From}\left(1\right)\right) \\ 2y+1=0\text{or}y+3=0 \\ y=-\frac{1}{2}\text{or}y=-3 \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

From the obtained values, the real solution of the equation is $y=-\frac{1}{2}\text{and}y=-3$.