What are the solutions to the following system of equations? y=x^2+12x+30 and 8x-y=10

Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:

8x-10 = x^2+12x+30

Group all terms together on the right side.  To do this, add -8x+10 to both sides.  Then 0 = x^2 +4x +40.  You must now solve this quadratic equation for x, if possible.  I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.

If you have a graphing calculator, please graph 8x-10  and  x^2+12x+30 on the same screen.  You will see two separate graphs that do NOT intersect.  This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!

The answer is