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t^2-3t-304=0
a = 1; b = -3; c = -304;
Δ = b2-4ac
Δ = -32-4·1·(-304)
Δ = 1225
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{1225}=35$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-35}{2*1}=\frac{-32}{2} =-16 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+35}{2*1}=\frac{38}{2} =19 $
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