How do you solve $7x^2 - 3x = 2$ using the quadratic formula?

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Answer 1 · 2016-03-27T12:51:04

Solution is $x=(3-sqrt(65))/14$ and $x=(3+sqrt(65))/14$

According to quadratic formula, the solution of quadratic equation $ax^2+bx+c=0$ is given by

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

As $7x^2-3x=2$ can be written in form $ax^2+bx+c=0$ as

$7x^2-3x-2=0$ , (i.e. $a=7$ , $b=-3$ and $c=-2$ ) and its solution will be given by

$x=(-(-3)+-sqrt((-3)^2-4xx7xx(-2)))/(2xx7)$ or

$x=(3+-sqrt(9+56))/14=(3+-sqrt(65))/14$

Hence, solution is $x=(3-sqrt(65))/14$ and $x=(3+sqrt(65))/14$

How do you solve 7x^2 - 3x = 27x^2 - 3x = 2 using the quadratic formula?