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t^2-9t-36=0
a = 1; b = -9; c = -36;
Δ = b2-4ac
Δ = -92-4·1·(-36)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*1}=\frac{-6}{2} =-3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*1}=\frac{24}{2} =12 $
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