Question

Can I find a tutor to help me wit this

Answer 1

`\begin{gathered} 1.3x^2-5x=-8 \\ \Rightarrow3x^2-5x+8=0 \\ D=b^2-4ac \\ D=(-5)^2-4(3)(8)=25-96=-71 \\ D=-71<0,\text{ No real solution} \\ It\text{ has two complex solution} \end{gathered}`

Next

`\begin{gathered} 2.\text{ }2x^2=6x-5 \\ \Rightarrow2x^2-6x+5=0 \\ D=(-6)^2-4(2)(5)=36-40=-4 \\ No\text{ real solution, It has two complex solution.} \end{gathered}`

Next

`\begin{gathered} 3.\text{ }12x=9x^2+4 \\ 9x^2-12x+4=0 \\ D=(-12)^2-4(9)(4)=144-144=0 \\ Real\text{ and equal roots} \end{gathered}`

Lastly,

`\begin{gathered} 4.\text{ }-x^2-10x=34 \\ x^2+10x+34=0 \\ D=(10)^2-4(1)(34)=100-136=-36 \\ No\text{ real solution, It has two complex solution} \end{gathered}`

Hence, the equations

`\begin{gathered} 3x^2-5x=-8,2x^2=6x-5,-x^2-10x=34\text{ have no real solution } \\ but\text{ have two complex solutions.} \end{gathered}`