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4t^2+16t-128=0
a = 4; b = 16; c = -128;
Δ = b2-4ac
Δ = 162-4·4·(-128)
Δ = 2304
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{2304}=48$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-48}{2*4}=\frac{-64}{8} =-8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+48}{2*4}=\frac{32}{8} =4 $
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