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20t+t^2-125=0
a = 1; b = 20; c = -125;
Δ = b2-4ac
Δ = 202-4·1·(-125)
Δ = 900
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{900}=30$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-30}{2*1}=\frac{-50}{2} =-25 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+30}{2*1}=\frac{10}{2} =5 $
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