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6=t+4/5t-3
We move all terms to the left:
6-(t+4/5t-3)=0
Domain of the equation: 5t-3)!=0We get rid of parentheses
t∈R
-t-4/5t+3+6=0
We multiply all the terms by the denominator
-t*5t+3*5t+6*5t-4=0
Wy multiply elements
-5t^2+15t+30t-4=0
We add all the numbers together, and all the variables
-5t^2+45t-4=0
a = -5; b = 45; c = -4;
Δ = b2-4ac
Δ = 452-4·(-5)·(-4)
Δ = 1945
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{-(45)-\sqrt{1945}}{2*-5}=\frac{-45-\sqrt{1945}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{1945}}{2*-5}=\frac{-45+\sqrt{1945}}{-10} $
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