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4t^2+t-39=0
a = 4; b = 1; c = -39;
Δ = b2-4ac
Δ = 12-4·4·(-39)
Δ = 625
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{625}=25$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-25}{2*4}=\frac{-26}{8} =-3+1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+25}{2*4}=\frac{24}{8} =3 $
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