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T=T2+5T+2
We move all terms to the left:
T-(T2+5T+2)=0
We add all the numbers together, and all the variables
-(+T^2+5T+2)+T=0
We get rid of parentheses
-T^2-5T+T-2=0
We add all the numbers together, and all the variables
-1T^2-4T-2=0
a = -1; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·(-1)·(-2)
Δ = 8
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}$$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*-1}=\frac{4-2\sqrt{2}}{-2} $$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*-1}=\frac{4+2\sqrt{2}}{-2} $
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