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2x^2+6x-1296=0
a = 2; b = 6; c = -1296;
Δ = b2-4ac
Δ = 62-4·2·(-1296)
Δ = 10404
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{10404}=102$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-102}{2*2}=\frac{-108}{4} =-27 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+102}{2*2}=\frac{96}{4} =24 $
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