Question

Find the real solutions of the equation${\left(5x+1\right)}^2+3=10$ by factoring.

Short answer

The real solution of the equation is $x=\frac{-1+\sqrt{7}}{5}\text{and}x=\frac{-1-\sqrt{7}}{5}$.

Step-by-step solution

Step 1. Factorized the given quadratic equation.

$$\begin{gathered} {\left(5x+1\right)}^2+3=10....\left(\text{Given}\right) \\ {\left(5x+1\right)}^2-7=0....\left(\text{Rearrangeandsimplify}\right) \\ {\left(5x+1\right)}^2-{\left(\sqrt{7}\right)}^2=0 \\ \left(5x+1-\sqrt{7}\right)\left(5x+1+\sqrt{7}\right)=0....\left(\text{Factorized}\right) \\ \left(5x+1-\sqrt{7}\right)\left(5x+1+\sqrt{7}\right)=0...\left(1\right) \end{gathered}$$

Step 2. Further simplifying (1).

$$\begin{gathered} \left(5x+1-\sqrt{7}\right)\left(5x+1+\sqrt{7}\right)=0....\left(\text{From}\left(1\right)\right) \\ 5x+1-\sqrt{7}=0\text{or}5x+1+\sqrt{7}=0 \\ x=\frac{-1+\sqrt{7}}{5}\text{or}x=\frac{-1-\sqrt{7}}{5} \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

From the obtained values, the real solution of the equation is$x=\frac{-1+\sqrt{7}}{5}\text{and}x=\frac{-1-\sqrt{7}}{5}$.