You can use completing the square method or the quadratic formula to find the roots and get the factor. Note the standard form of quadratic equation Ax² + Bx + C = 0.
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
s = √6 + 3 FIRST ROOT ANSWER
s = - √6 + 3 SECOND ROOT ANSWER
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
