How do you solve $x^2 – 3x = -6x – 1$ using the quadratic formula?

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Answer 1 · 2017-05-03T16:46:01

$color(red)(x=(-3+sqrt5)/2=-0.382$ to the nearest 3 decimal #
or

$color(red)(x=(-3-sqrt5)/2=-2.618$ to the nearest 3 decimal places

$x^2-3x=-6x-1$

$x^2+3x+1=0$

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

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

$x=(-3+-sqrt(9-4))/2$

$x=(-3+-sqrt5)/2$

$color(red)(x=(-3+sqrt5)/2=-0.382$ to the nearest 3 decimal places

or

$color(red)(x=(-3-sqrt5)/2=-2.618$ to the nearest 3 decimal places

How do you solve x^2 – 3x = -6x – 1x^2 – 3x = -6x – 1 using the quadratic formula?