How do you solve $x^{2} = 13 x - 42$ $x^2=13x-42$ ?

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Answer 1 · 2015-12-27T15:25:46

The roots of $x$ $x$ are $7$ $7$ and $6$ $6$

This equation may be written as

$x^{2} - 13 x + 42 = 0$ $x^2-13x+42=0$

or

$x^{2} - 7 x - 6 x + 42 = 0$ $x^2-7x-6x+42 =0$

or

$x (x - 7) - 6 (x - 7) = 0$ $x(x-7)-6(x-7)=0$

or

$(x - 7) (x - 6) = 0$ $(x-7)(x-6)=0$

So $x - 7 = 0$ $x-7=0$ or $x - 6 = 0$ $x-6=0$

therefore $x = 7$ $x=7$ or $6$ $6$
[answer]

How do you solve x2=13x−42x^2=13x-42?