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b^2-b-14=-3
We move all terms to the left:
b^2-b-14-(-3)=0
We add all the numbers together, and all the variables
b^2-1b-11=0
a = 1; b = -1; c = -11;
Δ = b2-4ac
Δ = -12-4·1·(-11)
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3\sqrt{5}}{2*1}=\frac{1-3\sqrt{5}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3\sqrt{5}}{2*1}=\frac{1+3\sqrt{5}}{2} $
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