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5(t)=-5t^2+30t+2
We move all terms to the left:
5(t)-(-5t^2+30t+2)=0
We get rid of parentheses
5t^2-30t+5t-2=0
We add all the numbers together, and all the variables
5t^2-25t-2=0
a = 5; b = -25; c = -2;
Δ = b2-4ac
Δ = -252-4·5·(-2)
Δ = 665
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{-(-25)-\sqrt{665}}{2*5}=\frac{25-\sqrt{665}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{665}}{2*5}=\frac{25+\sqrt{665}}{10} $
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