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2t^2-16t-80-40=0
We add all the numbers together, and all the variables
2t^2-16t-120=0
a = 2; b = -16; c = -120;
Δ = b2-4ac
Δ = -162-4·2·(-120)
Δ = 1216
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{19}}{2*2}=\frac{16-8\sqrt{19}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{19}}{2*2}=\frac{16+8\sqrt{19}}{4} $
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