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3x(x+1)(x-1)=36
We move all terms to the left:
3x(x+1)(x-1)-(36)=0
We use the square of the difference formula
x^2-1-36=0
We add all the numbers together, and all the variables
x^2-37=0
a = 1; b = 0; c = -37;
Δ = b2-4ac
Δ = 02-4·1·(-37)
Δ = 148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{37}}{2*1}=\frac{0-2\sqrt{37}}{2} =-\frac{2\sqrt{37}}{2} =-\sqrt{37} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{37}}{2*1}=\frac{0+2\sqrt{37}}{2} =\frac{2\sqrt{37}}{2} =\sqrt{37} $
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