Question

4x^2+x-5=-6x to the nearest tenth.

Answer 1

We want to solve

`\begin{gathered} 4x^2+x-5=-6x \\ bringing\text{ 6x to the other side, we have } \\ 4x^2+x+6x-5=0 \\ 4x^2+7x-5=0 \\ This\text{ is in the form of the quatric equation } \\ ax^2+bx+c=0 \\ So,\text{ it means } \\ a=4,b=7,c=-5 \end{gathered}`

Using the quadratic formula

`x=(-b\pm√(b^2-4ac))/(2a)`

We have

`\begin{gathered} x=(-b\pm√(b^2-4ac))/(2a) \\ x=(-7\pm√((-7)^2-4*4*-5))/(2*4) \\ x=(-7\pm√(49+80))/(8) \\ x=(-7\pm√(129))/(8) \end{gathered}`

So, either

`\begin{gathered} x=(-7+√(129))/(8) \\ x=(-7+11.35781669)/(8) \\ x=0.544727 \\ x=0.5 \end{gathered}`

Or

`\begin{gathered} x=(-7-√(129))/(8) \\ x=(-7-11.35781669)/(8) \\ x=-2.294727 \\ x=-2.3 \end{gathered}`

Hence the answer is x = 0.5 or -2.3 to the nearest tenth