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3t^2+21t-54=0
a = 3; b = 21; c = -54;
Δ = b2-4ac
Δ = 212-4·3·(-54)
Δ = 1089
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{1089}=33$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-33}{2*3}=\frac{-54}{6} =-9 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+33}{2*3}=\frac{12}{6} =2 $
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