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

May 15, 2015

${x}^{2} - x + 1 = 0$ (taking $- 1$ to the right and changing sign)
now your equation is in the form: $a {x}^{2} + b x + c = 0$ where the numerical coefficients are:
$a = 1$
$b = - 1$
$c = 1$
You can use them into the Quadratic Formula given as:
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
So:
${x}_{1 , 2} = \frac{1 \pm \sqrt{1 - 4 \left(1 \cdot 1\right)}}{2}$
${x}_{1 , 2} = \frac{1 \pm \sqrt{- 3}}{2}$
The negative argument of your square root cannot give you a real number.
The solutions of your equation will be two Complex Numbers .
You can write:
${x}_{1 , 2} = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2}$
${x}_{1 , 2} = \frac{1 \pm i \sqrt{3}}{2}$ where $i = \sqrt{- 1}$