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38=38-3t-5t^2
We move all terms to the left:
38-(38-3t-5t^2)=0
We get rid of parentheses
5t^2+3t-38+38=0
We add all the numbers together, and all the variables
5t^2+3t=0
a = 5; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·5·0
Δ = 9
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{9}=3$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*5}=\frac{-6}{10} =-3/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*5}=\frac{0}{10} =0 $
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