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