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By completing the square:

s² + 3 = 6s

s² - 6s = -3 Equation 1

Next get (B/2A)² :

(B/2A)² = {-6/2(1)}²

= (-3)²

= 9 ............ADD THIS TO BOTH SIDES OF EQUATION 1

We will have:

s² - 6s + 9 = -3 + 9

Left side of this equation is now a perfect square; thus factorable..

(s-3)² = 6

take the square root of both side to find your "s"

s-3 = √6

You can also use the quadratic formula: and you will get the same answer.

S = { -B +/- √ B² - 4AC } / 2A

S₁ = negative B PLUS the square root of the quantity B²-4AC all over 2A...for root 1.

S₂ = negative B MINUS the square root of the quantity B²-4AC all over 2A...for root 2