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10=263t-16t^2
We move all terms to the left:
10-(263t-16t^2)=0
We get rid of parentheses
16t^2-263t+10=0
a = 16; b = -263; c = +10;
Δ = b2-4ac
Δ = -2632-4·16·10
Δ = 68529
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-263)-\sqrt{68529}}{2*16}=\frac{263-\sqrt{68529}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-263)+\sqrt{68529}}{2*16}=\frac{263+\sqrt{68529}}{32} $
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