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5t^2-6t-61=0
a = 5; b = -6; c = -61;
Δ = b2-4ac
Δ = -62-4·5·(-61)
Δ = 1256
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1256}=\sqrt{4*314}=\sqrt{4}*\sqrt{314}=2\sqrt{314}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{314}}{2*5}=\frac{6-2\sqrt{314}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{314}}{2*5}=\frac{6+2\sqrt{314}}{10} $
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