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r^2-16r=43
We move all terms to the left:
r^2-16r-(43)=0
a = 1; b = -16; c = -43;
Δ = b2-4ac
Δ = -162-4·1·(-43)
Δ = 428
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{428}=\sqrt{4*107}=\sqrt{4}*\sqrt{107}=2\sqrt{107}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{107}}{2*1}=\frac{16-2\sqrt{107}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{107}}{2*1}=\frac{16+2\sqrt{107}}{2} $
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