b^2-23b+42=0

Solution for b^2-23b+42=0 equation:

b^2-23b+42=0

a = 1; b = -23; c = +42;
Δ = b2-4ac
Δ = -232-4·1·42
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-19}{2*1}=\frac{4}{2}
=2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+19}{2*1}=\frac{42}{2}
=21
$