b^2-16b+13=-7

Solution for b^2-16b+13=-7 equation:

b^2-16b+13=-7

We move all terms to the left:

b^2-16b+13-(-7)=0

We add all the numbers together, and all the variables

b^2-16b+20=0

a = 1; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·1·20
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{11}}{2*1}=\frac{16-4\sqrt{11}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{11}}{2*1}=\frac{16+4\sqrt{11}}{2}
$