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8t-3/2t+5=-19
We move all terms to the left:
8t-3/2t+5-(-19)=0
Domain of the equation: 2t!=0We add all the numbers together, and all the variables
t!=0/2
t!=0
t∈R
8t-3/2t+24=0
We multiply all the terms by the denominator
8t*2t+24*2t-3=0
Wy multiply elements
16t^2+48t-3=0
a = 16; b = 48; c = -3;
Δ = b2-4ac
Δ = 482-4·16·(-3)
Δ = 2496
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{39}}{2*16}=\frac{-48-8\sqrt{39}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{39}}{2*16}=\frac{-48+8\sqrt{39}}{32} $
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