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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)
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)
Answer:
13.3
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