Question

What the answer to this equation be?

Answer 1

If you expand the square, you have

$\dfrac{1}{4}x^2-3x=7x+8-10x$

Simplify the right hand side:

$\dfrac{1}{4}x^2-3x=-3x+8$

Cancel the -3x appearing on both sides:

$\dfrac{1}{4}x^2=8$

Multiply both sides by 4:

$x^2=32$

Consider the square root of both terms (using the doubles sign):

$x=\pm\sqrt{32}=\pm4\sqrt{2}$