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3x^2+24x-1152=0
a = 3; b = 24; c = -1152;
Δ = b2-4ac
Δ = 242-4·3·(-1152)
Δ = 14400
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{14400}=120$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-120}{2*3}=\frac{-144}{6} =-24 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+120}{2*3}=\frac{96}{6} =16 $
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