Answer
No Real Solution
Work Step by Step
$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$
$w^2 = 3(w-1)$
Distribute the 3 to $(w-1)$
$w^2 = 3w-3$
Subtract 3w from both sides
$w^2-3w = 3w-3-3w$
$w^2-3w = -3$
Add 3 to both sides
$w^2-3w+3 = -3+3$
$w^2-3w+3=0$
Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$
$1w^2-3w+3$
$a = 1, b = -3, c = 3$
$x = \frac{-(-3) \pm \sqrt {(-3)^2- 4(1 \times 3)}}{2(1)}$
$x = \frac{3 \pm \sqrt {9- 12}}{2}$
$x= \frac{3 \pm \sqrt {-3}}{2}$
You can never square root a negative number, so there is no real solution possible
No Real Solution
