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5t^2-18t-9=0
a = 5; b = -18; c = -9;
Δ = b2-4ac
Δ = -182-4·5·(-9)
Δ = 504
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{14}}{2*5}=\frac{18-6\sqrt{14}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{14}}{2*5}=\frac{18+6\sqrt{14}}{10} $
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