Answer
$x = -1.4 \quad \ \text{or}\ \ \ x= 0 $
Work Step by Step
Divide both sides of the equation by $5$ so that the square term has a coefficient of one. Then, proceed with completing the square of the equation by dividing the coefficient of the second term term by $2$ and squaring it and adding the result on both sides.
$$
\begin{aligned}
5x^2+7x& = 0 \\
x^2+1.4x &= 0\\
x^2+1.4x+\left(\frac{1.4}{2}\right)^2 & =\left(\frac{1.4}{2}\right)^2 \\
x^2+1.4x +0.7^2& = 0.7^2\\
(x+0.7)^2&= 0.7^2.
\end{aligned}
$$ Take the square root: $$
\begin{array}{rl}
(x+0.7)^2&= 0.7^2 \\
x+0.7 & = \pm \sqrt{0.7^2} \\
x+0.7 & = \pm 0.7 \\\\
x+0.7=0.7 &\ \text{or}\ \ \ x+0.7=-0.7\\
x = 0.7-0.7 & \ \text{or}\ \ \ x = -0.7-0.7 \\
x = 0 & \ \textbf{or}\ \ \ x = -1.4 \\
\end{array}
$$
Check $$\begin{aligned}
5(0)^2+7(0)& \stackrel{?}{=} 0 \\
0& = 0\checkmark
\end{aligned}$$ $$\begin{aligned}
5(-1.4)^2+7(-1.4)& \stackrel{?}{=} 0 \\
0& = 0\checkmark.
\end{aligned}$$ The solution is:
$$x = -1.4 \quad \ \text{or}\ \ \ x= 0 $$
