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-3a^2+36a+39=0
a = -3; b = 36; c = +39;
Δ = b2-4ac
Δ = 362-4·(-3)·39
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1764}=42$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-42}{2*-3}=\frac{-78}{-6} =+13 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+42}{2*-3}=\frac{6}{-6} =-1 $
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