Question

what are the solutions to the following system of equations?
y = x² + 12x + 30
8x - y = 10
(-4, -2) and (2, 5)
(-2, -4) and (2, 5)
(-2, -4) and (5, 2)
no real solutions

Explanation:

Step1: Isolate y from linear equation

Rearrange $8x - y = 10$ to solve for y:
$y = 8x - 10$

Step2: Substitute into quadratic equation

Replace y in $y = x^2 + 12x + 30$:
$8x - 10 = x^2 + 12x + 30$

Step3: Rearrange to standard quadratic form

Bring all terms to one side:
$x^2 + 12x + 30 - 8x + 10 = 0$
$x^2 + 4x + 40 = 0$

Step4: Calculate discriminant

Use discriminant formula $\Delta = b^2 - 4ac$ for $ax^2+bx+c=0$:
$\Delta = 4^2 - 4(1)(40) = 16 - 160 = -144$

Step5: Analyze discriminant

Since $\Delta < 0$, there are no real roots.

Answer:

No real solutions