11b+b^2=42

Solution for 11b+b^2=42 equation:

11b+b^2=42

We move all terms to the left:

11b+b^2-(42)=0

a = 1; b = 11; c = -42;
Δ = b2-4ac
Δ = 112-4·1·(-42)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-17}{2*1}=\frac{-28}{2}
=-14
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+17}{2*1}=\frac{6}{2}
=3
$