13b+b^2+36=0

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Solution for 13b+b^2+36=0 equation: 



13b+b^2+36=0

a = 1; b = 13; c = +36;
Δ = b2-4ac
Δ = 132-4·1·36
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*1}=\frac{-18}{2}
=-9
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*1}=\frac{-8}{2}
=-4
$

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