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3t^2+26t-77=0
a = 3; b = 26; c = -77;
Δ = b2-4ac
Δ = 262-4·3·(-77)
Δ = 1600
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{1600}=40$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-40}{2*3}=\frac{-66}{6} =-11 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+40}{2*3}=\frac{14}{6} =2+1/3 $
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