2q^2+19q+45=0

Solution for 2q^2+19q+45=0 equation:

2q^2+19q+45=0

a = 2; b = 19; c = +45;
Δ = b2-4ac
Δ = 192-4·2·45
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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