2x^2-23x+73=x^2-13x+48

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Solution for 2x^2-23x+73=x^2-13x+48 equation: 



2x^2-23x+73=x^2-13x+48

We move all terms to the left:

2x^2-23x+73-(x^2-13x+48)=0

We get rid of parentheses

2x^2-x^2-23x+13x-48+73=0

We add all the numbers together, and all the variables

x^2-10x+25=0

a = 1; b = -10; c = +25;
Δ = b2-4ac
Δ = -102-4·1·25
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{10}{2}=5$

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