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r^2-14r=14
We move all terms to the left:
r^2-14r-(14)=0
a = 1; b = -14; c = -14;
Δ = b2-4ac
Δ = -142-4·1·(-14)
Δ = 252
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{7}}{2*1}=\frac{14-6\sqrt{7}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{7}}{2*1}=\frac{14+6\sqrt{7}}{2} $
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