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4r^2=4r+1
We move all terms to the left:
4r^2-(4r+1)=0
We get rid of parentheses
4r^2-4r-1=0
a = 4; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·4·(-1)
Δ = 32
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{2}}{2*4}=\frac{4-4\sqrt{2}}{8} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{2}}{2*4}=\frac{4+4\sqrt{2}}{8} $
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