The quadratic formula states that the solutions to the equation
are (if they exist)
![x_(1,2) = (-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2019/formulas/mathematics/high-school/346vkdz3mbazhifbkh7b7bxuok6ohul4vh.png)
In your case, the quadratic equation is identified by the choice
, and this leads to the solutions
![x_(1,2) = (4\pm√(16+32))/(2) = (4\pm√(48))/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4lz0bx3ieg738uxhf04vdncyz1ij8x6ai4.png)
Since
, we have
![√(48) = √(16\cdot 3) = √(16)\codt √(3) = 4√(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/f6q6pw74qy73y3sqmorsilj8eorgzbpqze.png)
So, the expression for the solutions becomes
![(4\pm4√(3))/(2) = 2\pm2√(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/9aut0j7ewn5o77oc2w9rz7s78n0lle78yk.png)
So, the two solutions are
and
![2 - 2√(3)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/q0ffra8vf75657jg6v4tu2qlrcptdqcwc4.png)