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5t^2+12t-27=0
a = 5; b = 12; c = -27;
Δ = b2-4ac
Δ = 122-4·5·(-27)
Δ = 684
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{684}=\sqrt{36*19}=\sqrt{36}*\sqrt{19}=6\sqrt{19}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{19}}{2*5}=\frac{-12-6\sqrt{19}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{19}}{2*5}=\frac{-12+6\sqrt{19}}{10} $
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