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-16(t^2-20t+36)=0
We multiply parentheses
-16t^2+320t-576=0
a = -16; b = 320; c = -576;
Δ = b2-4ac
Δ = 3202-4·(-16)·(-576)
Δ = 65536
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{65536}=256$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-256}{2*-16}=\frac{-576}{-32} =+18 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+256}{2*-16}=\frac{-64}{-32} =+2 $
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