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(R)=(340+4R)(75-3R)
We move all terms to the left:
(R)-((340+4R)(75-3R))=0
We add all the numbers together, and all the variables
R-((4R+340)(-3R+75))=0
We multiply parentheses ..
-((-12R^2+300R-1020R+25500))+R=0
We calculate terms in parentheses: -((-12R^2+300R-1020R+25500)), so:We get rid of parentheses
(-12R^2+300R-1020R+25500)
We get rid of parentheses
-12R^2+300R-1020R+25500
We add all the numbers together, and all the variables
-12R^2-720R+25500
Back to the equation:
-(-12R^2-720R+25500)
12R^2+720R+R-25500=0
We add all the numbers together, and all the variables
12R^2+721R-25500=0
a = 12; b = 721; c = -25500;
Δ = b2-4ac
Δ = 7212-4·12·(-25500)
Δ = 1743841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(721)-\sqrt{1743841}}{2*12}=\frac{-721-\sqrt{1743841}}{24} $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(721)+\sqrt{1743841}}{2*12}=\frac{-721+\sqrt{1743841}}{24} $
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