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4.9t^2-15t-40t=0
We add all the numbers together, and all the variables
4.9t^2-55t=0
a = 4.9; b = -55; c = 0;
Δ = b2-4ac
Δ = -552-4·4.9·0
Δ = 3025
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{3025}=55$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-55}{2*4.9}=\frac{0}{9.8} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+55}{2*4.9}=\frac{110}{9.8} =11+2.2/9.8 $
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