Evaluate: 6y^2 = 7y - 5

Question:

Evaluate: {eq}6y^2= 7y - 5 {/eq}

An equation which is of the form {eq}ax^2+bx+c=0 {/eq} where {eq}a, b \text{ and } c {/eq} are constants is called a quadratic equation in {eq}x {/eq}. It has two (either same or distinct) roots. The roots can be solved by using the quadratic formula which states:

$$x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$

The given equation is:

$$6y^2= 7y - 5 \\ \text{Subtract 7y and add 5 on both sides}, \\ 6y^2-7y+5=0$$

Comparing this with {eq}ay^2+by+c=0 {/eq}, we get:

$$a=6 \\ b=-7 \\ c=5$$

We substitute all these values in the quadratic formula:

\begin{align} \color{red}{ x} &\color{red}{ =} \color{red}{ \dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\[0.4cm] y&= \frac{-(-7) \pm \sqrt{(-7)^{2}-4 \cdot 6 \cdot 5}}{2 \cdot 6} \\[0.4cm] &=\frac{7 \pm \sqrt{49-120}}{12} \\[0.4cm] &=\frac{7\pm \sqrt{-71}}{12} \\[0.4cm] &=\frac{7\pm i\sqrt{71}}{12} & [ \because \sqrt{-1}=i ] \\[0.4cm] \color{blue}{ y} &\color{blue}{ =} \color{blue}{ \boxed{\mathbf{\frac{7}{12}+i \frac{\sqrt{71}}{12}; \,\,\,\, \frac{7}{12}-i \frac{\sqrt{71}}{12}}}} \end{align}