How do you solve $2x^2 + 7x = 5$ ?

Question

5821 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-31T02:17:32

$x=(-7+sqrt89)/4;$ $x=(-7-sqrt89)/4$

Solve $2x^2+7x=5$

Subtract $5$ from both sides.

$2x^2+7x-5=0$

This is a quadratic equation in the form $ax^2+7x-5$ , where $a=2$ , $b=7$ , $c=-5$ .

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

Plug in the values from the quadratic formula.

$x=(-7+-sqrt(7^2-4*2*-5))/(2*2)$

$x=(-7+-sqrt(49+40))/4$

$x=(-7+-sqrt(89))/(4)$

$89$ is a prime number so it cannot be factored further.

$x=(-7+sqrt89)/4;$ $x=(-7-sqrt89)/4$

How do you solve 2x^2 + 7x = 52x^2 + 7x = 5?