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-16-2t=3/2t+9
We move all terms to the left:
-16-2t-(3/2t+9)=0
Domain of the equation: 2t+9)!=0We get rid of parentheses
t∈R
-2t-3/2t-9-16=0
We multiply all the terms by the denominator
-2t*2t-9*2t-16*2t-3=0
Wy multiply elements
-4t^2-18t-32t-3=0
We add all the numbers together, and all the variables
-4t^2-50t-3=0
a = -4; b = -50; c = -3;
Δ = b2-4ac
Δ = -502-4·(-4)·(-3)
Δ = 2452
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2452}=\sqrt{4*613}=\sqrt{4}*\sqrt{613}=2\sqrt{613}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{613}}{2*-4}=\frac{50-2\sqrt{613}}{-8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{613}}{2*-4}=\frac{50+2\sqrt{613}}{-8} $
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