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2t^2+21t+40=0
a = 2; b = 21; c = +40;
Δ = b2-4ac
Δ = 212-4·2·40
Δ = 121
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{121}=11$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*2}=\frac{-32}{4} =-8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*2}=\frac{-10}{4} =-2+1/2 $
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