Question

$6x^2+7x-10=0$

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{6}{x}^2+7x-10=0 \\ \mathrm{On}\mathrm{c}\text{omparing it with}a{x}^2+bx+c=0,\text{we get:} \\ a=6,b=7\mathrm{and}c=-10 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}[7^2-4\times 6\times (-10)] \\ \text{=}49+240 \\ \text{= 289 > 0} \\ \mathrm{Thus}\text{, the roots of the equation are real.} \\ \mathrm{Roots}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{given}\mathrm{by}: \\ \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-7+\sqrt{289}}{2\times 6}=\frac{-7+\sqrt{289}}{12}=\frac{-7+17}{12}=\frac{10}{12}=\frac{5}{6} \\ \beta \text{=}\frac{-b-\sqrt{D}}{2a}=\frac{-7-\sqrt{289}}{2\times 6}=\frac{-7-\sqrt{289}}{12}=\frac{-7-17}{12}=\frac{-24}{12}=-2 \\ \text{Hence, the roots of the equation are}\frac{5}{6}\text{and}-2\text{.} \end{gathered}$$