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

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Answer 1 · 2015-07-20T07:34:03

$x=(3+sqrt(19)i)/2, (3-sqrt(19)i)/2$

$x^2-3x=-7$

Get all terms on the left side.

$x^2-3x+7=0$ is a quadratic equation in the form $ax^2+bx+c=0$ , where $a=1, b=-3, and c=7$ .

Quadratic Equation

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

Substitute known values into the formula.

$x=(-(-3)+-sqrt(-3^2-4*1*7))/(2*1)$ =

$x=(3+-sqrt(9-28))/2$

$x=(3+-sqrt(-19))/2$

$x=(3+-sqrt(19)i)/2$

Solve for $x$ .

$x=(3+sqrt(19)i)/2$

$x=(3-sqrt(19)i)/2$

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