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5t^2-10t-25=0
a = 5; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·5·(-25)
Δ = 600
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{6}}{2*5}=\frac{10-10\sqrt{6}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{6}}{2*5}=\frac{10+10\sqrt{6}}{10} $
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