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t^2+11t-4=0
a = 1; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·1·(-4)
Δ = 137
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{137}}{2*1}=\frac{-11-\sqrt{137}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{137}}{2*1}=\frac{-11+\sqrt{137}}{2} $
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