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5t^2-25t+2=0
a = 5; b = -25; c = +2;
Δ = b2-4ac
Δ = -252-4·5·2
Δ = 585
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3\sqrt{65}}{2*5}=\frac{25-3\sqrt{65}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3\sqrt{65}}{2*5}=\frac{25+3\sqrt{65}}{10} $
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