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255-21t-16t^2=0
a = -16; b = -21; c = +255;
Δ = b2-4ac
Δ = -212-4·(-16)·255
Δ = 16761
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{-(-21)-\sqrt{16761}}{2*-16}=\frac{21-\sqrt{16761}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{16761}}{2*-16}=\frac{21+\sqrt{16761}}{-32} $
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