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16t^2-30t-4=0
a = 16; b = -30; c = -4;
Δ = b2-4ac
Δ = -302-4·16·(-4)
Δ = 1156
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{1156}=34$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-34}{2*16}=\frac{-4}{32} =-1/8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+34}{2*16}=\frac{64}{32} =2 $
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