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=4T+3-T2
We move all terms to the left:
-(4T+3-T2)=0
We add all the numbers together, and all the variables
-(+4T-1T^2+3)=0
We get rid of parentheses
1T^2-4T-3=0
We add all the numbers together, and all the variables
T^2-4T-3=0
a = 1; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·1·(-3)
Δ = 28
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}$$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{7}}{2*1}=\frac{4-2\sqrt{7}}{2} $$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{7}}{2*1}=\frac{4+2\sqrt{7}}{2} $
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