2x^2+19=1-20x

Solution for 2x^2+19=1-20x equation:

2x^2+19=1-20x

We move all terms to the left:

2x^2+19-(1-20x)=0

We add all the numbers together, and all the variables

2x^2-(-20x+1)+19=0

We get rid of parentheses

2x^2+20x-1+19=0

We add all the numbers together, and all the variables

2x^2+20x+18=0

a = 2; b = 20; c = +18;
Δ = b2-4ac
Δ = 202-4·2·18
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*2}=\frac{-36}{4}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*2}=\frac{-4}{4}
=-1
$