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10=5t^2-3t
We move all terms to the left:
10-(5t^2-3t)=0
We get rid of parentheses
-5t^2+3t+10=0
a = -5; b = 3; c = +10;
Δ = b2-4ac
Δ = 32-4·(-5)·10
Δ = 209
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{-(3)-\sqrt{209}}{2*-5}=\frac{-3-\sqrt{209}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{209}}{2*-5}=\frac{-3+\sqrt{209}}{-10} $
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