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3x^2+12x-1440=0
a = 3; b = 12; c = -1440;
Δ = b2-4ac
Δ = 122-4·3·(-1440)
Δ = 17424
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{17424}=132$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-132}{2*3}=\frac{-144}{6} =-24 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+132}{2*3}=\frac{120}{6} =20 $
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