b^2-10b-23=0

Solution for b^2-10b-23=0 equation:

b^2-10b-23=0

a = 1; b = -10; c = -23;
Δ = b2-4ac
Δ = -102-4·1·(-23)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8\sqrt{3}}{2*1}=\frac{10-8\sqrt{3}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8\sqrt{3}}{2*1}=\frac{10+8\sqrt{3}}{2}
$