If you expand the square, you have
[tex]\dfrac{1}{4}x^2-3x=7x+8-10x[/tex]
Simplify the right hand side:
[tex]\dfrac{1}{4}x^2-3x=-3x+8[/tex]
Cancel the -3x appearing on both sides:
[tex]\dfrac{1}{4}x^2=8[/tex]
Multiply both sides by 4:
[tex]x^2=32[/tex]
Consider the square root of both terms (using the doubles sign):
[tex]x=\pm\sqrt{32}=\pm4\sqrt{2}[/tex]
