b^2+9b-15=0

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Solution for b^2+9b-15=0 equation: 



b^2+9b-15=0

a = 1; b = 9; c = -15;
Δ = b2-4ac
Δ = 92-4·1·(-15)
Δ = 141
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{-(9)-\sqrt{141}}{2*1}=\frac{-9-\sqrt{141}}{2}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{141}}{2*1}=\frac{-9+\sqrt{141}}{2}
$

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