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5t^2-25t+18=0
a = 5; b = -25; c = +18;
Δ = b2-4ac
Δ = -252-4·5·18
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{-(-25)-\sqrt{265}}{2*5}=\frac{25-\sqrt{265}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{265}}{2*5}=\frac{25+\sqrt{265}}{10} $
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