1 Answer

Brian Deragon · Answer 1 · 2023-06-15T10:59:43+0000

we have the equation

3y^2 - 6y = 1

equate to zero

3y^2 - 6y - 1=0

a=3

b=-6

c=-1

substitute the given values in the formula

$y=\frac{-(-6)\pm\sqrt[]{-6^2-4(3)(-1)}}{2(3)}$
$y=\frac{6\pm\sqrt[]{48}}{6}$

simplify

$y=\frac{6\pm4\sqrt[]{3}}{6}$

the solutions for y are

$y=1+\frac{2\sqrt[]{3}}{3}$

and

$y=1-\frac{2\sqrt[]{3}}{3}$

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