How do you solve $92 - 42 x = 2 x^{2}$ $92- 42x = 2x^2$ ?

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How do you solve $92 - 42 x = 2 x^{2}$ $92- 42x = 2x^2$ ?

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Answer 1 · 2018-05-02T07:39:31

$x = 2 or x = - 23$ $x = 2 or x = -23$

Explanation:

Notice that it is a quadratic equation because there is an $x^{2}$ $x^2$ term. The first step is to move all the terms to one side to make the equation equal to $0$ $0$

$2 x^{2} + 42 x - 92 = 0 \leftarrow \div 2$ $2x^2 +42x-92=0" "larr div 2$

$x^{2} + 21 x - 46 = 0 \leftarrow$ $x^2 +21x-46=0" "larr$ find factors if possible

$(x + 23) (x - 2) = 0$ $(x+23)(x-2)=0$

Either factor could be equal to $0$ $0$

$x + 23 = 0 \to x = - 23$ $x+23 = 0 " "rarr x =-23$

$x - 2 = 0 \to x = 2$ $x-2=0" "rarr x = 2$

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How do you solve $92 - 42 x = 2 x^{2}$ $92- 42x = 2x^2$ ?

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How do you solve 92−42x=2x292- 42x = 2x^2?

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How do you solve $92 - 42 x = 2 x^{2}$ $92- 42x = 2x^2$ ?