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5t^2+20t-41=0
a = 5; b = 20; c = -41;
Δ = b2-4ac
Δ = 202-4·5·(-41)
Δ = 1220
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1220}=\sqrt{4*305}=\sqrt{4}*\sqrt{305}=2\sqrt{305}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{305}}{2*5}=\frac{-20-2\sqrt{305}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{305}}{2*5}=\frac{-20+2\sqrt{305}}{10} $
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