To solve quadratics, the most correct and effective way, in my opinion, is the quadratic formula:
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Plugging it in, we get:
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So the final answer is:
and
Pascals triangle to the 6th:
1 x^0
1 1 x^1
1 2 1 x^2
1 3 3 1 x^3
1 4 6 4 1 x^4
1 5 10 10 5 1 x^5
1 6 15 20 15 6 1 x^6<span>
</span>the problem is to the 6th power so your going to use the 6th row of pascals triangle (don't count the first row). these numbers represent the coefficients of the variables
1(d-5y)^6 + 6(d-5y)^5 + 15(d-5y)^4 + 20(d-5y)^3 + 15(d-5y)^2 + 6(d-5y) + 1
then simplify
We can readily know the x^2-4x+4=3y then use it to replace the same function in the first equation which refers to the 3y+y^2-6y=0
y^2-3y=0
y(y-3)=0
y1=0 -----------y2=3
Seven hundred and and eighty thousand
