How do you solve $x^{2} + 6 = 5 x$ $x^2+6=5x$ ?

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Answer 1 · 2016-06-12T12:09:20

$x = 3$ $x=3$ or $x = 2$ $x=2$

$x^{2} + 6 = 5 \cdot x$ $x^2+6 = 5*x$

$x^{2} - 5 \cdot x + 6 = 0$ $x^2 -5*x +6=0$

It can be written as two factors in the form:

$(x + a) \cdot (x + b) = 0$ $(x+a)*(x+b) =0$ .

Where a, b are two integers

such that :
$a + b = - 5$ $a + b = -5$

and

$a \cdot b = 6$ $a* b = 6$

Thus, $a = - 2$ $a=-2$ and $b = - 3$ $b= -3$

Substituting the values of a and b we have:

$x^{2} - 5 \cdot x + 6 = 0$ $x^2 -5*x +6 =0$

$(x - 3) \cdot (x - 2) = 0$ $(x-3)*(x-2)=0$

$x - 3 = 0$ $x-3=0$
Or
$x - 2 = 0$ $x-2=0$

so,
$x = 3$ $x=3$
or
$x = 2$ $x=2$

How do you solve x^2+6=5xx2+6=5xx^2+6=5x?