5w^2+19w+18=0

Solution for 5w^2+19w+18=0 equation:

5w^2+19w+18=0

a = 5; b = 19; c = +18;
Δ = b2-4ac
Δ = 192-4·5·18
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*5}=\frac{-20}{10}
=-2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*5}=\frac{-18}{10}
=-1+4/5
$