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5t^2+5t-67=0
a = 5; b = 5; c = -67;
Δ = b2-4ac
Δ = 52-4·5·(-67)
Δ = 1365
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{1365}}{2*5}=\frac{-5-\sqrt{1365}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1365}}{2*5}=\frac{-5+\sqrt{1365}}{10} $
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