x^2-4x+38=-x^2+16x-12

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Solution for x^2-4x+38=-x^2+16x-12 equation: 



x^2-4x+38=-x^2+16x-12

We move all terms to the left:

x^2-4x+38-(-x^2+16x-12)=0

We get rid of parentheses

x^2+x^2-16x-4x+12+38=0

We add all the numbers together, and all the variables

2x^2-20x+50=0

a = 2; b = -20; c = +50;
Δ = b2-4ac
Δ = -202-4·2·50
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{20}{4}=5$

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