$0=4{x}^{2}+8x-5$
Swap the sides of the equation$4{x}^{2}+8x-5=0$
Write $8x$ as a difference$4{x}^{2}+10x-2x-5=0$
Factor out $2x$ from the expression$2x \times \left( 2x+5 \right)-2x-5=0$
Factor out the negative sign from the expression$2x \times \left( 2x+5 \right)-\left( 2x+5 \right)=0$
Factor out $2x+5$ from the expression$\left( 2x+5 \right) \times \left( 2x-1 \right)=0$
When the product of factors equals $0$, at least one factor is $0$$\begin{array} { l }2x+5=0,\\2x-1=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=-\frac{ 5 }{ 2 },\\2x-1=0\end{array}$
Solve the equation for $x$$\begin{array} { l }x=-\frac{ 5 }{ 2 },\\x=\frac{ 1 }{ 2 }\end{array}$
The equation has $2$ solutions$\begin{align*}&\begin{array} { l }x_1=-\frac{ 5 }{ 2 },& x_2=\frac{ 1 }{ 2 }\end{array} \\&\begin{array} { l }x_1=-2.5,& x_2=0.5\end{array}\end{align*}$