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t-(2t+5)-5t(1-2t)=2(3+4t)-3(t-4)
We move all terms to the left:
t-(2t+5)-5t(1-2t)-(2(3+4t)-3(t-4))=0
We add all the numbers together, and all the variables
t-(2t+5)-5t(-2t+1)-(2(4t+3)-3(t-4))=0
We multiply parentheses
10t^2+t-(2t+5)-5t-(2(4t+3)-3(t-4))=0
We get rid of parentheses
10t^2+t-2t-5t-(2(4t+3)-3(t-4))-5=0
We calculate terms in parentheses: -(2(4t+3)-3(t-4)), so:We add all the numbers together, and all the variables
2(4t+3)-3(t-4)
We multiply parentheses
8t-3t+6+12
We add all the numbers together, and all the variables
5t+18
Back to the equation:
-(5t+18)
10t^2-6t-(5t+18)-5=0
We get rid of parentheses
10t^2-6t-5t-18-5=0
We add all the numbers together, and all the variables
10t^2-11t-23=0
a = 10; b = -11; c = -23;
Δ = b2-4ac
Δ = -112-4·10·(-23)
Δ = 1041
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{-(-11)-\sqrt{1041}}{2*10}=\frac{11-\sqrt{1041}}{20} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{1041}}{2*10}=\frac{11+\sqrt{1041}}{20} $
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