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4t^2-64t=0
a = 4; b = -64; c = 0;
Δ = b2-4ac
Δ = -642-4·4·0
Δ = 4096
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{4096}=64$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-64}{2*4}=\frac{0}{8} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+64}{2*4}=\frac{128}{8} =16 $
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