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36x^2+12x+1=18
We move all terms to the left:
36x^2+12x+1-(18)=0
We add all the numbers together, and all the variables
36x^2+12x-17=0
a = 36; b = 12; c = -17;
Δ = b2-4ac
Δ = 122-4·36·(-17)
Δ = 2592
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36\sqrt{2}}{2*36}=\frac{-12-36\sqrt{2}}{72} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36\sqrt{2}}{2*36}=\frac{-12+36\sqrt{2}}{72} $
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