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5t^2+29t-6=0
a = 5; b = 29; c = -6;
Δ = b2-4ac
Δ = 292-4·5·(-6)
Δ = 961
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-31}{2*5}=\frac{-60}{10} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+31}{2*5}=\frac{2}{10} =1/5 $
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