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3x^2+12x-495=0
a = 3; b = 12; c = -495;
Δ = b2-4ac
Δ = 122-4·3·(-495)
Δ = 6084
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{6084}=78$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-78}{2*3}=\frac{-90}{6} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+78}{2*3}=\frac{66}{6} =11 $
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