Answer
$x=-\dfrac{7}{2}\pm\dfrac{\sqrt{51}}{2}$
Work Step by Step
$2x^{2}+14x-1=0$
Take the $-1$ to the right side of the equation:
$2x^{2}+14x=1$
Take out common factor $2$ from the left side of the equation:
$2(x^{2}+7x)=1$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $2\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular equation, $b=7$
$2\Big[x^{2}+7x+\Big(\dfrac{7}{2}\Big)^{2}\Big]=1+2\Big(\dfrac{7}{2}\Big)^{2}$
$2\Big(x^{2}+7x+\dfrac{49}{4}\Big)=1+\dfrac{49}{2}$
Factor the polynomial inside the parentheses, which is a perfect square trinomial:
$2\Big(x+\dfrac{7}{2}\Big)^{2}=\dfrac{51}{2}$
Take the $2$ to divide the right side of the equation:
$\Big(x+\dfrac{7}{2}\Big)^{2}=\dfrac{51}{4}$
Take the square root of both sides of the equation:
$\sqrt{\Big(x+\dfrac{7}{2}\Big)^{2}}=\pm\sqrt{\dfrac{51}{4}}$
$x+\dfrac{7}{2}=\pm\dfrac{\sqrt{51}}{2}$
Solve for $x$:
$x=-\dfrac{7}{2}\pm\dfrac{\sqrt{51}}{2}$
