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4t^2+9t+1=0
a = 4; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·4·1
Δ = 65
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{-(9)-\sqrt{65}}{2*4}=\frac{-9-\sqrt{65}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{65}}{2*4}=\frac{-9+\sqrt{65}}{8} $
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