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4t^2-2t-30=0
a = 4; b = -2; c = -30;
Δ = b2-4ac
Δ = -22-4·4·(-30)
Δ = 484
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{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*4}=\frac{-20}{8} =-2+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*4}=\frac{24}{8} =3 $
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