Question

Solve the following quadratic equation using the quadratic formula. (Formula to use is in the picture attached) - 2y^2 - 6y = 3

Answer 1

`\begin{gathered} -2y^2-6y=3 \\ -2y^2-6y-3=0 \\ \end{gathered}`

then, a = -2, b = -6 and c = -3

`\begin{gathered} y_(1,2)=(-\left(-6\right)\pm√(\left(-6\right)^2-4\left(-2\right)\left(-3\right)))/(2\left(-2\right)) \\ y_(1,\: 2)=\frac{6\pm\sqrt[]{36-24}}{-4} \\ y_(1,\: 2)=-\frac{6\pm\sqrt[]{12}}{4} \\ \end{gathered}`

so, solve for y1 and y2

`y_1=-\frac{6+\sqrt[]{12}}{4}=-\frac{6+2\sqrt[]{3}}{4}=-\frac{2(3+\sqrt[]{3})}{4}=-\frac{3+\sqrt[]{3}}{2}`
`y_2=-\frac{6-\sqrt[]{12}}{4}=-\frac{6-2\sqrt[]{3}}{4}=-\frac{2(3-\sqrt[]{3})}{4}=-\frac{3-\sqrt[]{3}}{2}`