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5t^2+23t+10=0
a = 5; b = 23; c = +10;
Δ = b2-4ac
Δ = 232-4·5·10
Δ = 329
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{-(23)-\sqrt{329}}{2*5}=\frac{-23-\sqrt{329}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{329}}{2*5}=\frac{-23+\sqrt{329}}{10} $
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