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2t^2-16t-18=0
a = 2; b = -16; c = -18;
Δ = b2-4ac
Δ = -162-4·2·(-18)
Δ = 400
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{400}=20$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*2}=\frac{-4}{4} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*2}=\frac{36}{4} =9 $
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