5x(x-4)=x^2+8x-49

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Solution for 5x(x-4)=x^2+8x-49 equation: 



5x(x-4)=x^2+8x-49

We move all terms to the left:

5x(x-4)-(x^2+8x-49)=0

We multiply parentheses

5x^2-20x-(x^2+8x-49)=0

We get rid of parentheses

5x^2-x^2-20x-8x+49=0

We add all the numbers together, and all the variables

4x^2-28x+49=0

a = 4; b = -28; c = +49;
Δ = b2-4ac
Δ = -282-4·4·49
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{28}{8}=3+1/2$

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