How do you solve 2x^2+x-1=9?

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How do you solve 2x^2+x-1=9?

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Answer 1 · 2018-05-08T14:08:31

$x = - \frac{5}{2} or x = 2$ $x=-5/2 or x=2$

Explanation:

$2 x^{2} + x - 1 = 9 \Rightarrow 2 x^{2} + x - 10 = 0$ $2x^2+x-1=9 => 2x^2+x-10=0$

$(2 x + 5) (x - 2) = 0$ $(2x+5)(x-2)=0$

$2 x + 5 = 0 or x - 2 = 0$ $2x+5=0 or x-2=0$

$2 x = - 5 or x = 2$ $2x=-5 or x=2$

$x = - \frac{5}{2} or x = 2$ $x=-5/2 or x=2$

Answer 2 · 2018-05-08T14:13:16

$x = - \frac{5}{2}, 2$ $x=-5/2 , 2$

Explanation:

let,
$2 x^{2} + x - 1 - 9 = 0$ $2x^2+x-1-9=0$
$2 x^{2} - 4 x + 5 x - 10$ $2x^2-4x+5x-10$ =0 (we get this by factorizing)
$2 x (x - 2) + 5 (x - 2)$ $2x(x-2)+5(x-2)$ =0
$(2 x + 5) (x - 2) = 0$ $(2x+5)(x-2)=0$
$therefore$ $x= -5/2,2$

Answer 3 · 2018-05-08T15:39:17

$x=-5/2$ and $x=2$

Explanation:

Set equal to 0:

$2x^2+x-1=9$

$rArr 2x^2+x-10=0$

Factorise:

$2x^2-4x+5x-10$

$rArr 2x(x-2)+5(x-2)$

$rArr (2x+5)(x-2)$

Set equal to 0:

$2x+5=0$

$rArr 2x=-5$

$rArr x=-5/2$

$x-2=0$

$rArrx=2$

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How do you solve 2x^2+x-1=9?

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