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3t^2-18t-48=0
a = 3; b = -18; c = -48;
Δ = b2-4ac
Δ = -182-4·3·(-48)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{900}=30$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-30}{2*3}=\frac{-12}{6} =-2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+30}{2*3}=\frac{48}{6} =8 $
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