No solution for the quadratic equation because of the √-20
2
![2x^2+7x-10 = x+5 2x^2+7x-15 y2x^2+6x-15](https://tex.z-dn.net/?f=2x%5E2%2B7x-10%20%3D%20x%2B5%0A%0A2x%5E2%2B7x-15%0A%0Ay2x%5E2%2B6x-15%20)
Suggesting that you want this in standard form, in terms of quadratic equations, you would technically follow a process similar if not almost exactly like the 2 - step equation method with the exception of separating the (x)s and the equations to find x and then plug it in and what-not.
With that being said you would subtract 5 in (x+5) from said 5 in the second equation and -10 in the first equation in order to get 2x^2+7x-15, you would continue to do the same for the x by subtracting it from both ends making the 7x a 6x because there is a 1 at the beginning of each x if there is no number that is shown already. Which finally gives you the equation (y= 2x^2+6x-15)
I believe your answer would be c