How do you solve $2x^2+5x=0$ using the quadratic formula?

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How do you solve $2x^2+5x=0$ using the quadratic formula?

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Answer 1 · 2016-08-13T21:00:37

$x={0,-5/2}$

Explanation:

$ax^2+bx+c=0$

$2x^2+5x=0$

$a=2" ; "b=5" ; "c=0$

$Delta=sqrt(b^2-4*a*c)$

$Delta=sqrt(5^2-4*2*0)$

$Delta=sqrt 5^2$

$Delta=±5$

$x_("1,2")=(-b±Delta)/(2*a)$

$x_1=(-5-5)/(2*2)=-10/4=-5/2$

$x_2=(-5+5)/(2*2)=0$

$x={0,-5/2}$

$"Or simply"$

$2x^2+5x=0$

$2cancel(x^2)=-5cancel(x)$

$2x=-5$

$x=-5/2$

Answer 2 · 2016-08-13T21:59:12

$x=-5/2, 0$

Explanation:

$2x^2+5x=0$ is a quadratic equation in the form $ax^2+bx+c$ , where $a=2$ , $5=b$ , and $c=0$ .

Quadratic formula

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Plug the given values into the formula.

$x=(-5+-sqrt((-5)^2-4*2*0))/(2*2)$

Simplify.

$x=(-5+-sqrt(25-0))/4$

Simplify.

$x=(-5+-sqrt 25)/4$

Simplify.

$x=(-5+-5)/4$

Solve for $x_1$ and $x_2$ .

$color(blue)(x_1=(-5+5)/4)$

$color(blue)(x_1=0/4)$

$color(blue)(x_1=0)$

$color(red)(x_2=(-5-5)/4)$

$color(red)(x_2=-10/4)$

$color(red)(x_2=-5/2)$

$color(purple)(x=-5/2, 0)$

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How do you solve $2x^2+5x=0$ using the quadratic formula?

2 Answers

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