In order to solve the system of equations, let's equate both values of y and solve for x using the quadratic formula:
[tex]\begin{gathered} 2x^2-5x+10=2x+5\\ \\ 2x^2-5x-2x+10-5=0\\ \\ 2x^2-7x+5=0\\ \\ a=2,b=-7,c=5\\ \\ x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\ \\ x=\frac{7±\sqrt{49-40}}{4}\\ \\ x_1=\frac{7+3}{4}=\frac{10}{4}=\frac{5}{2}\\ \\ x_2=\frac{7-3}{4}=\frac{4}{4}=1 \end{gathered}[/tex]
Now, let's calculate the values of y for each value of x:
[tex]\begin{gathered} x=\frac{5}{2}:\\ \\ y=2x+5=5+5=10\\ \\ \\ \\ x=1:\\ \\ y=2x+5=2+5=7 \end{gathered}[/tex]
Therefore the solutions are (1, 7) and (5/2, 10).
Correct option: second one.
