b2-6b-53=10

Solution for b2-6b-53=10 equation:

b2-6b-53=10

We move all terms to the left:

b2-6b-53-(10)=0

We add all the numbers together, and all the variables

b^2-6b-63=0

a = 1; b = -6; c = -63;
Δ = b2-4ac
Δ = -62-4·1·(-63)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12\sqrt{2}}{2*1}=\frac{6-12\sqrt{2}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12\sqrt{2}}{2*1}=\frac{6+12\sqrt{2}}{2}
$