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4.9t^2+20t+12=0
a = 4.9; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·4.9·12
Δ = 164.8
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{-(20)-\sqrt{164.8}}{2*4.9}=\frac{-20-\sqrt{164.8}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{164.8}}{2*4.9}=\frac{-20+\sqrt{164.8}}{9.8} $
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