How do you solve $20 - 9 x + x^{2} = 0$ $20-9x+x^2=0$ ?

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How do you solve $20 - 9 x + x^{2} = 0$ $20-9x+x^2=0$ ?

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Answer 1 · 2016-08-31T16:28:37

x=5 or x=3

Explanation:

The first step is to factor:
$x^{2} - 9 x + 20 = 0$ $x^2 - 9x + 20 = 0$
$(x - 5) (x - 4) = 0$ $(x-5) (x-4) = 0$

The second step is to set each factor to zero:
$x = 5$ $x=5$ or $x = 3$ $x=3$

Answer 2 · 2016-08-31T16:32:30

$x = 4$ $x=4$ or $x = 5$ $x=5$

Explanation:

The first step when solving for solutions is to factor the equation:
$x^{2} - 9 x + 20 = 0$ $x^2-9x+20=0$ , which when factored will be: $(x - 4) (x - 5) = 0$ $(x-4)(x-5)=0$

The next step is to solve each for zero:
$(x - 4) (x - 5) = 0$ $(x-4)(x-5)=0$
$x = 4$ $x=4$ or
$x = 5$ $x=5$

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How do you solve $20 - 9 x + x^{2} = 0$ $20-9x+x^2=0$ ?

2 Answers

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