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5t^2-6t-11=0
a = 5; b = -6; c = -11;
Δ = b2-4ac
Δ = -62-4·5·(-11)
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-16}{2*5}=\frac{-10}{10} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+16}{2*5}=\frac{22}{10} =2+1/5 $
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