b^2-11b=-19

Solution for b^2-11b=-19 equation:

b^2-11b=-19

We move all terms to the left:

b^2-11b-(-19)=0

We add all the numbers together, and all the variables

b^2-11b+19=0

a = 1; b = -11; c = +19;
Δ = b2-4ac
Δ = -112-4·1·19
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3\sqrt{5}}{2*1}=\frac{11-3\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3\sqrt{5}}{2*1}=\frac{11+3\sqrt{5}}{2}
$