3x^2+5x+20=2x^2-3x+4

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Solution for 3x^2+5x+20=2x^2-3x+4 equation: 



3x^2+5x+20=2x^2-3x+4

We move all terms to the left:

3x^2+5x+20-(2x^2-3x+4)=0

We get rid of parentheses

3x^2-2x^2+5x+3x-4+20=0

We add all the numbers together, and all the variables

x^2+8x+16=0

a = 1; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·1·16
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-8}{2}=-4$

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