b2+6b-19=-10

Solution for b2+6b-19=-10 equation:

b2+6b-19=-10

We move all terms to the left:

b2+6b-19-(-10)=0

We add all the numbers together, and all the variables

b^2+6b-9=0

a = 1; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·1·(-9)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{2}}{2*1}=\frac{-6-6\sqrt{2}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{2}}{2*1}=\frac{-6+6\sqrt{2}}{2}
$