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t2=5t+7
We move all terms to the left:
t2-(5t+7)=0
We add all the numbers together, and all the variables
t^2-(5t+7)=0
We get rid of parentheses
t^2-5t-7=0
a = 1; b = -5; c = -7;
Δ = b2-4ac
Δ = -52-4·1·(-7)
Δ = 53
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{-(-5)-\sqrt{53}}{2*1}=\frac{5-\sqrt{53}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{53}}{2*1}=\frac{5+\sqrt{53}}{2} $
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