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5t^2-20t+11=0
a = 5; b = -20; c = +11;
Δ = b2-4ac
Δ = -202-4·5·11
Δ = 180
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{5}}{2*5}=\frac{20-6\sqrt{5}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{5}}{2*5}=\frac{20+6\sqrt{5}}{10} $
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