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36x^2-420x-132=0
a = 36; b = -420; c = -132;
Δ = b2-4ac
Δ = -4202-4·36·(-132)
Δ = 195408
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{195408}=\sqrt{144*1357}=\sqrt{144}*\sqrt{1357}=12\sqrt{1357}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-420)-12\sqrt{1357}}{2*36}=\frac{420-12\sqrt{1357}}{72} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-420)+12\sqrt{1357}}{2*36}=\frac{420+12\sqrt{1357}}{72} $
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