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t^2-11t-12=0
a = 1; b = -11; c = -12;
Δ = b2-4ac
Δ = -112-4·1·(-12)
Δ = 169
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{169}=13$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*1}=\frac{-2}{2} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*1}=\frac{24}{2} =12 $
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