b^2-28b+15=0

Solution for b^2-28b+15=0 equation:

b^2-28b+15=0

a = 1; b = -28; c = +15;
Δ = b2-4ac
Δ = -282-4·1·15
Δ = 724
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:

$\sqrt{\Delta}=\sqrt{724}=\sqrt{4*181}=\sqrt{4}*\sqrt{181}=2\sqrt{181}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{181}}{2*1}=\frac{28-2\sqrt{181}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{181}}{2*1}=\frac{28+2\sqrt{181}}{2}
$