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4t+16t^2-20=0
a = 16; b = 4; c = -20;
Δ = b2-4ac
Δ = 42-4·16·(-20)
Δ = 1296
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{1296}=36$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*16}=\frac{-40}{32} =-1+1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*16}=\frac{32}{32} =1 $
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