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b^2+5b+12=70
We move all terms to the left:
b^2+5b+12-(70)=0
We add all the numbers together, and all the variables
b^2+5b-58=0
a = 1; b = 5; c = -58;
Δ = b2-4ac
Δ = 52-4·1·(-58)
Δ = 257
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{257}}{2*1}=\frac{-5-\sqrt{257}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{257}}{2*1}=\frac{-5+\sqrt{257}}{2} $
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