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5t^2+12t-65=0
a = 5; b = 12; c = -65;
Δ = b2-4ac
Δ = 122-4·5·(-65)
Δ = 1444
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1444}=38$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-38}{2*5}=\frac{-50}{10} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+38}{2*5}=\frac{26}{10} =2+3/5 $
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