#### Answer

$x=\dfrac{\sqrt{7}-1}{5}$
$x=-\dfrac{1+\sqrt{7}}{5}$

#### Work Step by Step

$(5x+1)^{2}+3=10$
We need to solve this by factoring, so let's take the $10$ to the left side of the equation to substract:
$(5x+1)^{2}+3-10=0$
Simplify:
$(5x+1)^{2}-7=0$
Remember the difference of two squares factoring formula:
$a^{2}-b^{2}=(a-b)(a+b)$
In this equation $a=(5x+1)^{2}$ and $b=7$, so we can factor it and we get:
$[(5x+1)-\sqrt{7}][(5x+1)+\sqrt{7}]=0$
Then, we set each factor equal to $0$ and solve them as individual equations. Let the $[(5x+1)-\sqrt{7}]=0$ be equation number 1 and $[(5x+1)+\sqrt{7}]=0$ be equation number 2:
Solving for $x$ in equation number 1:
$(5x+1)-\sqrt{7}=0$
$5x=\sqrt{7}-1$
$x=\dfrac{\sqrt{7}-1}{5}$
Solving for $x$ in equation number 2:
$(5x+1)+\sqrt{7}=0$
$5x=-1-\sqrt{7}$
$x=-\dfrac{1+\sqrt{7}}{5}$