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b^2-12b=-34
We move all terms to the left:
b^2-12b-(-34)=0
We add all the numbers together, and all the variables
b^2-12b+34=0
a = 1; b = -12; c = +34;
Δ = b2-4ac
Δ = -122-4·1·34
Δ = 8
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{2}}{2*1}=\frac{12-2\sqrt{2}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{2}}{2*1}=\frac{12+2\sqrt{2}}{2} $
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