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t.4(t+1/4)=3
We move all terms to the left:
t.4(t+1/4)-(3)=0
We add all the numbers together, and all the variables
t.4(+t+1/4)-3=0
We multiply parentheses
t^2+t^2-3=0
We add all the numbers together, and all the variables
2t^2-3=0
a = 2; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·2·(-3)
Δ = 24
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*2}=\frac{0-2\sqrt{6}}{4} =-\frac{2\sqrt{6}}{4} =-\frac{\sqrt{6}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*2}=\frac{0+2\sqrt{6}}{4} =\frac{2\sqrt{6}}{4} =\frac{\sqrt{6}}{2} $
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