b^2-24b+40=0

Solution for b^2-24b+40=0 equation:

b^2-24b+40=0

a = 1; b = -24; c = +40;
Δ = b2-4ac
Δ = -242-4·1·40
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{26}}{2*1}=\frac{24-4\sqrt{26}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{26}}{2*1}=\frac{24+4\sqrt{26}}{2}
$