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4t^2-29t+7=0
a = 4; b = -29; c = +7;
Δ = b2-4ac
Δ = -292-4·4·7
Δ = 729
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{729}=27$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-27}{2*4}=\frac{2}{8} =1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+27}{2*4}=\frac{56}{8} =7 $
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