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b^2-9b-3b-4=0
We add all the numbers together, and all the variables
b^2-12b-4=0
a = 1; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·1·(-4)
Δ = 160
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{10}}{2*1}=\frac{12-4\sqrt{10}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*1}=\frac{12+4\sqrt{10}}{2} $
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