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10t^2-5t-10=0
a = 10; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·10·(-10)
Δ = 425
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{17}}{2*10}=\frac{5-5\sqrt{17}}{20} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{17}}{2*10}=\frac{5+5\sqrt{17}}{20} $
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