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(R)=-4R^2+40R+135
We move all terms to the left:
(R)-(-4R^2+40R+135)=0
We get rid of parentheses
4R^2-40R+R-135=0
We add all the numbers together, and all the variables
4R^2-39R-135=0
a = 4; b = -39; c = -135;
Δ = b2-4ac
Δ = -392-4·4·(-135)
Δ = 3681
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}$$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3681}=\sqrt{9*409}=\sqrt{9}*\sqrt{409}=3\sqrt{409}$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-3\sqrt{409}}{2*4}=\frac{39-3\sqrt{409}}{8} $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+3\sqrt{409}}{2*4}=\frac{39+3\sqrt{409}}{8} $
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