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6t^2+5t-10=0
a = 6; b = 5; c = -10;
Δ = b2-4ac
Δ = 52-4·6·(-10)
Δ = 265
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{265}}{2*6}=\frac{-5-\sqrt{265}}{12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{265}}{2*6}=\frac{-5+\sqrt{265}}{12} $
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