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t^2+65t-12=0
a = 1; b = 65; c = -12;
Δ = b2-4ac
Δ = 652-4·1·(-12)
Δ = 4273
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{-(65)-\sqrt{4273}}{2*1}=\frac{-65-\sqrt{4273}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+\sqrt{4273}}{2*1}=\frac{-65+\sqrt{4273}}{2} $
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