Question

The roots of y in the equation y^2-2y=-9/2 are +-

Answer 1

`\bf y^2-2y=-\cfrac{9}{2}\implies y^2-2y+\cfrac{9}{2}=0\\\\ -------------------------------\\\\ \qquad \qquad \textit{quadratic formula}\\\\ \begin{array}{lccclll} &{{ 1}}x^2&{{ -2}}x&{{ +\cfrac{9}{2}}}&=0\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad y= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}`


`\bf y=\cfrac{-(-2)\pm\sqrt{(-2)^2-4(1)\left( (9)/(2) \right)}}{2(1)}\implies y=\cfrac{2\pm√(4-18)}{2} \\\\\\ y=1\pm√(-14)\implies y=1\pm√(-1\cdot 14)\implies y=1\pm√(-1)\cdot √(14) \\\\\\ y=1\pm i√(14)`

Answer 2

`y= \frac{2+\isqrt{14}}{2}`

`y= \frac{2-\isqrt{14}}{2}`

Explanation:

`y^2 - 2y = (-9)/(2)`

add 9/2 on both sides

`y^2 - 2y + (9)/(2)=0`

Apply quadratic formula and solve for y

a= 1, b=-2 and c= 9/2

`y= (-b+-√(b^2-4ac))/(2a)`

`y= (2+-√((-2)^2-4*1*9/2))/(2(1))`

`y= (2+-i√(14))/(2)`

WE cannot simplify it further

`y= (2+i√(14))/(2)`

`y= (2-i√(14))/(2)`