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5t^2-8t-12=0
a = 5; b = -8; c = -12;
Δ = b2-4ac
Δ = -82-4·5·(-12)
Δ = 304
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{19}}{2*5}=\frac{8-4\sqrt{19}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{19}}{2*5}=\frac{8+4\sqrt{19}}{10} $
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