#### Answer

The solutions are $x=-5\pm2\sqrt{7}$

#### Work Step by Step

$x^{2}+10x-3=0$
Take the $3$ to the right side of the equation:
$x^{2}+10x=3$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=10$
$x^{2}+10x+\Big(\dfrac{10}{2}\Big)^{2}=3+\Big(\dfrac{10}{2}\Big)^{2}$
$x^{2}+10x+25=3+25$
$x^{2}+10x+25=28$
Factor the left side of the equation, which is a perfect square trinomial:
$(x+5)^{2}=28$
Take the square root of both sides:
$\sqrt{(x+5)^{2}}=\pm\sqrt{28}$
$x+5=\pm2\sqrt{7}$
Solve for $x$:
$x=-5\pm2\sqrt{7}$