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

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Answer 1 · 2015-07-17T04:04:49

$x=5+sqrt(23), x=5-sqrt(23)$

$x^2-3x=7x-2$

Get all of the terms on the left side.

$x^2-3x-7x+2=0$ =

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

The equation is now in the form of a quadratic equation $ax^2+bx+c$ , where $a=1,$ $b=-10,$ and $c=2$ .

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

Substitute the values for $a, b, and c$ into the formula.

$x=(-(-10)+-sqrt(-10^2-4*1*2))/(2*1)$ =

$x=(10+-sqrt(100-8))/(2)$ =

$x=(10+-sqrt(92))/2$

Simplify $sqrt(92)$ .

$sqrt(92)=sqrt(2xx2xx23)$ =

$sqrt(92)=2sqrt23$

Substitute $2sqrt(23)$ for $sqrt(92)$ .

$x=(10+-2sqrt(23))/2$

Simplify.

$x=5+-sqrt(23)$

Solve for $x$ .

$x=5+sqrt(23)$

$x=5-sqrt(23)$

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