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5/12q+1/2q=25-3
We move all terms to the left:
5/12q+1/2q-(25-3)=0
Domain of the equation: 12q!=0
q!=0/12
q!=0
q∈R
Domain of the equation: 2q!=0We add all the numbers together, and all the variables
q!=0/2
q!=0
q∈R
5/12q+1/2q-22=0
We calculate fractions
10q/24q^2+12q/24q^2-22=0
We multiply all the terms by the denominator
10q+12q-22*24q^2=0
We add all the numbers together, and all the variables
22q-22*24q^2=0
Wy multiply elements
-528q^2+22q=0
a = -528; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-528)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-528}=\frac{-44}{-1056} =1/24 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-528}=\frac{0}{-1056} =0 $
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