Answer:
[tex]\large\boxed{x=\dfrac{1}{2}-\sqrt2\ or\ x=\dfrac{1}{2}+\sqrt2}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula of}\ ax^2+bx+c=0:\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]\text{We have:}\\\\4x^2-7=4x\qquad\text{subtract}\ 4x\ \text{from both sides}\\\\4x^2-4x-7=0\\\\a=4,\ b=-4,\ c=-7\\\\b^2-4ac=(-4)^2-4(4)(-7)=16+112=128\\\\\sqrt{b^2-4ac}=\sqrt{128}=\sqrt{64\cdot2}=\sqrt{64}\cdot\sqrt2=8\sqrt2\\\\x=\dfrac{-(-4)\pm8\sqrt2}{(2)(4)}=\dfrac{4\pm8\sqrt2}{8}\qquad\text{simplify by 4}\\\\x=\dfrac{1\pm2\sqrt2}{2}\to x=\dfrac{1}{2}\pm\sqrt2[/tex]
