b^2-12b-28=0

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Solution for b^2-12b-28=0 equation:



b^2-12b-28=0
a = 1; b = -12; c = -28;
Δ = b2-4ac
Δ = -122-4·1·(-28)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-16}{2*1}=\frac{-4}{2} =-2 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+16}{2*1}=\frac{28}{2} =14 $

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