a^2+18a-40=0

Solution for a^2+18a-40=0 equation:

a^2+18a-40=0

a = 1; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·1·(-40)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*1}=\frac{-40}{2}
=-20
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*1}=\frac{4}{2}
=2
$