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3/5t^2=4
We move all terms to the left:
3/5t^2-(4)=0
Domain of the equation: 5t^2!=0We multiply all the terms by the denominator
t^2!=0/5
t^2!=√0
t!=0
t∈R
-4*5t^2+3=0
Wy multiply elements
-20t^2+3=0
a = -20; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-20)·3
Δ = 240
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-20}=\frac{0-4\sqrt{15}}{-40} =-\frac{4\sqrt{15}}{-40} =-\frac{\sqrt{15}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-20}=\frac{0+4\sqrt{15}}{-40} =\frac{4\sqrt{15}}{-40} =\frac{\sqrt{15}}{-10} $
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