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5t^2-40t-1200=0
a = 5; b = -40; c = -1200;
Δ = b2-4ac
Δ = -402-4·5·(-1200)
Δ = 25600
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{25600}=160$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-160}{2*5}=\frac{-120}{10} =-12 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+160}{2*5}=\frac{200}{10} =20 $
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