2b^2+18b=40

Solution for 2b^2+18b=40 equation:

2b^2+18b=40

We move all terms to the left:

2b^2+18b-(40)=0

a = 2; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·2·(-40)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{161}}{2*2}=\frac{-18-2\sqrt{161}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{161}}{2*2}=\frac{-18+2\sqrt{161}}{4}
$