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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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*-4}=\frac{-24}{-8} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*-4}=\frac{20}{-8} =-2+1/2 $
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