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4.9t^2-15t+8=0
a = 4.9; b = -15; c = +8;
Δ = b2-4ac
Δ = -152-4·4.9·8
Δ = 68.2
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{-(-15)-\sqrt{68.2}}{2*4.9}=\frac{15-\sqrt{68.2}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{68.2}}{2*4.9}=\frac{15+\sqrt{68.2}}{9.8} $
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