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3t^2+3t-2500=0
a = 3; b = 3; c = -2500;
Δ = b2-4ac
Δ = 32-4·3·(-2500)
Δ = 30009
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{-(3)-\sqrt{30009}}{2*3}=\frac{-3-\sqrt{30009}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{30009}}{2*3}=\frac{-3+\sqrt{30009}}{6} $
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