Answer:
The two points of intersection are approximately (-2.56155, -7.3693) and (1.56155, 17.3693)
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Explanation:
Apply substitution and get everything to one side
[tex]y = 6x+8\\\\x^2+7x+4 = 6x+8\\\\x^2+7x+4-6x-8 = 0\\\\x^2+x-4 = 0\\\\[/tex]
Now apply the quadratic formula with a = 1, b = 1, c = -4
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-1\pm\sqrt{(1)^2-4(1)(-4)}}{2(1)}\\\\x = \frac{-1\pm\sqrt{1+16}}{2}\\\\x = \frac{-1\pm\sqrt{17}}{2}\\\\x = \frac{-1+\sqrt{17}}{2} \ \text{ or } \ x = \frac{-1-\sqrt{17}}{2}\\\\x \approx 1.56155\ \text{ or } \ x \approx -2.56155\\\\[/tex]
Each approximate x value is then plugged into either original equation to find its corresponding paired y value.
Let's plug the first x value found.
[tex]y = 6x+8\\\\y \approx 6(1.56155)+8\\\\y \approx 17.3693\\\\[/tex]
Therefore [tex](x,y) \approx (1.56155, 17.3693)[/tex] is one approximate point of intersection.
Repeat for the other x value
[tex]y = 6x+8\\\\y = 6(-2.56155)+8\\\\y = -7.3693\\\\[/tex]
The other point of intersection is roughly located at [tex](x,y) \approx (-2.56155, -7.3693)[/tex]
