Use the elimination method to solve the linear-quadratic system y=x2+2x−4 , y=−x−6 .

To use the elimination method, we can set the two equations equal to each other:

x^2 + 2x - 4 = -x - 6

Now, let's rearrange the equation to set it equal to zero:

x^2 + 3x - 2 = 0

Now, we have a quadratic equation that we can solve using quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a
x = (-3 ± sqrt(3^2 - 4*1*(-2))) / 2*1
x = (-3 ± sqrt(9 + 8)) / 2
x = (-3 ± sqrt(17)) / 2

Therefore, the solutions for x are:

x = (-3 + sqrt(17)) / 2
x = (-3 - sqrt(17)) / 2

Now, we can plug these values of x back into one of the original equations to find the corresponding values of y:

For x = (-3 + sqrt(17)) / 2:

y = (-3 + sqrt(17))^2 + 2(-3 + sqrt(17)) - 4
y = (9 - 6sqrt(17) + 17) - 6 + 2sqrt(17) - 4
y = 16 - 4
y = 12

For x = (-3 - sqrt(17)) / 2:

y = (-3 - sqrt(17))^2 + 2(-3 - sqrt(17)) - 4
y = (9 + 6sqrt(17) + 17) - 6 - 2sqrt(17) - 4
y = 26 - 4
y = 22

Therefore, the solutions to the system of equations are:

x = (-3 + sqrt(17)) / 2, y = 12
x = (-3 - sqrt(17)) / 2, y = 22