How do you solve $f (x) = x^{2} - x + 3$ $f(x)=x^2-x+3$ using the quadratic formula?

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Answer 1 · 2017-07-19T11:25:19

See a solution process below:

For $f (x) = a x^{2} + b x + c$ $f(x) = ax^2 + bx + c$ , the values of $x$ $x$ which are the solutions to the equation for $0$ $0$ are given by:

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

Substituting $1$ $1$ for $a$ $a$ ; $- 1$ $-1$ for $b$ $b$ and $3$ $3$ for $c$ $c$ gives:

$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - (4 \cdot 1 \cdot 3)}}{2 \cdot 1}$ $x = (-(-1) +- sqrt((-1)^2 - (4 * 1 * 3)))/(2 * 1)$

$x = \frac{1 \pm \sqrt{1 - 12}}{2}$ $x = (1 +- sqrt(1 - 12))/2$

$x = \frac{1 \pm \sqrt{- 11}}{2}$ $x = (1 +- sqrt(-11))/2$

How do you solve f(x)=x2−x+3 f(x)=x^2-x+3 using the quadratic formula?