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-3x^2-6x+36=0
a = -3; b = -6; c = +36;
Δ = b2-4ac
Δ = -62-4·(-3)·36
Δ = 468
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{13}}{2*-3}=\frac{6-6\sqrt{13}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{13}}{2*-3}=\frac{6+6\sqrt{13}}{-6} $
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