3x2+5x+20=16+x-x2

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Solution for 3x2+5x+20=16+x-x2 equation: 



3x^2+5x+20=16+x-x2

We move all terms to the left:

3x^2+5x+20-(16+x-x2)=0

We add all the numbers together, and all the variables

3x^2-(+x-1x^2+16)+5x+20=0

We get rid of parentheses

3x^2+1x^2-x+5x-16+20=0

We add all the numbers together, and all the variables

4x^2+4x+4=0

a = 4; b = 4; c = +4;
Δ = b2-4ac
Δ = 42-4·4·4
Δ = -48
Delta is less than zero, so there is no solution for the equation

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