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2t^2-11t-13=0
a = 2; b = -11; c = -13;
Δ = b2-4ac
Δ = -112-4·2·(-13)
Δ = 225
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{225}=15$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-15}{2*2}=\frac{-4}{4} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+15}{2*2}=\frac{26}{4} =6+1/2 $
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