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0=-4.9t^2-3t+50
We move all terms to the left:
0-(-4.9t^2-3t+50)=0
We add all the numbers together, and all the variables
-(-4.9t^2-3t+50)=0
We get rid of parentheses
4.9t^2+3t-50=0
a = 4.9; b = 3; c = -50;
Δ = b2-4ac
Δ = 32-4·4.9·(-50)
Δ = 989
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{989}=\sqrt{1*989}=\sqrt{1}*\sqrt{989}=1\sqrt{989}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1\sqrt{989}}{2*4.9}=\frac{-3-1\sqrt{989}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1\sqrt{989}}{2*4.9}=\frac{-3+1\sqrt{989}}{9.8} $
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