Answer:
[tex]\large\boxed{x=\dfrac{5-\sqrt5}{4},\ x=\dfrac{5+\sqrt5}{4}}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then the equation has no real solution}\\\\\text{if}\ b^2-4ac=0,\ \text{then the equation has one solution:}\ x=\dfrac{-b}{2a}\\\\\text{if}\ b^2-4ac,\ ,\ \text{then the equation has two solutions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\==========================================[/tex]
[tex]\text{We have the equation:}\ 4x^2-10x+5=0\\\\a=4,\ b=-10,\ c=5\\\\b^2-4ac=(-10)^2-4(4)(5)=100-80=20>0\\\\x=\dfrac{-(-10)\pm\sqrt{20}}{2(4)}=\dfrac{10\pm\sqrt{4\cdot5}}{8}=\dfrac{10\pm\sqrt4\cdot\sqrt5}{8}=\dfrac{10\pm2\sqrt5}{8}\\\\=\dfrac{2(5\pm\sqrt5)}{8}=\dfrac{5\pm\sqrt5}{4}[/tex]
