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6x^2-36x-3=0
a = 6; b = -36; c = -3;
Δ = b2-4ac
Δ = -362-4·6·(-3)
Δ = 1368
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1368}=\sqrt{36*38}=\sqrt{36}*\sqrt{38}=6\sqrt{38}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{38}}{2*6}=\frac{36-6\sqrt{38}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{38}}{2*6}=\frac{36+6\sqrt{38}}{12} $
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