2x^2+20x+45=-3

Solution for 2x^2+20x+45=-3 equation:

2x^2+20x+45=-3

We move all terms to the left:

2x^2+20x+45-(-3)=0

We add all the numbers together, and all the variables

2x^2+20x+48=0

a = 2; b = 20; c = +48;
Δ = b2-4ac
Δ = 202-4·2·48
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*2}=\frac{-16}{4}
=-4
$