How do you solve $x^{2} - x = - 1$ $x^2-x= -1$ using the quadratic formula?

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Answer 1 · 2016-07-26T15:50:38

$x = \frac{1 \pm i \sqrt{3}}{2}$ $x=(1+-isqrt(3))/2$

The quadratic formula for a general quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

For the equation:

$x^{2} - x = - 1$ $x^2 - x = -1$

or $x^{2} - x + 1 = 0$ $x^2 -x + 1=0$

you get

$a = 1; b = - 1 and c = 1$ $a=1; b=-1 and c = 1$

by substituting these values in the quadratic formula:

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$ $x=(-(-1)+-sqrt((-1)^2-4*1*1))/(2*1)$

$x = \frac{1 \pm \sqrt{1 - 4}}{2}$ $x= (1+-sqrt(1-4))/2$

or $x = \frac{1 \pm \sqrt{- 3}}{2}$ $x=(1+-sqrt(-3))/2$

or $x = \frac{1 \pm i \sqrt{3}}{2}$ $x=(1+-isqrt(3))/2$

How do you solve x2−x=−1x^2-x= -1 using the quadratic formula?