Question #de83c

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Question #de83c

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Answer 1 · 2017-11-02T13:23:57

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

Explanation:

$2 x^{2} - 25 - 5 x = 0$ $2x^2-25-5x=0$
$2 x^{2} - 5 x - 25 = 0$ $2x^2-5x-25=0$
$2 x^{2} - 10 x + 5 x - 25 = 0$ $2x^2-10x+5x-25=0$
$2 x (x - 5) + 5 (x - 5) = 0$ $2x(x-5)+5(x-5)=0$
$(2 x + 5) (x - 5) = 0$ $(2x+5)(x-5)=0$
$(2 x + 5) = 0$ $(2x+5)=0$ or $x - 5 = 0$ $x-5=0$
$2 x = - 5$ $2x=-5$ or $x = 5$ $x=5$
$x = \frac{- 5}{2}$ $x=(-5)/2$ or $x = 5$ $x=5$

Answer 2 · 2017-11-02T14:12:56

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

Explanation:

The quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Given equation is $2 x^{2} - 5 x - 25$ $2x^2-5x-25$

Here $a = 2$ $a=2$ , $b = - 5$ $b=-5$ and $c = - 25$ $c=-25$

Put these values in the quadratic formula.

$x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 \times (2) \times (- 25)}}{2 \times 2}$ $x=(-(-5)+-sqrt((-5)^2-4 xx (2) xx (-25)))/(2 xx 2)$

$x = \frac{5 \pm \sqrt{25 + 200}}{4}$ $x=(5+-sqrt(25+200))/4$

$x = \frac{5 \pm 15}{4}$ $x=(5+-15)/4$

So $x = \frac{5 + 15}{4}$ $x=(5+15)/4$ or $x = \frac{5 - 15}{4}$ $x=(5-15)/4$

$x = \frac{20}{4}$ $x=20/4$ or $x = \frac{- 10}{4}$ $x=(-10)/4$

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

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Question #de83c

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