How do I solve this quadratic and linear equation using substitution? I know that the answer is (6,-2) (-9,-47) But I cannot figure out the steps to get the answers.
y=3x-20
y=x^2+34
Thank you for your help.

Question

How do I solve this quadratic and linear equation using substitution? I know that the answer is (6,-2) (-9,-47) But I cannot figure out the steps to get the answers. 
y=3x-20
y=x^2+34
 Thank you for your help.

anonymous · Answer

write 
$$3x-20=x^2+34$$and solve for $x$

anonymous · Answer

$$ y = 3x - 20 $$
substitute this y in the second equation:
$$ 3x - 20 = x^2 + 34 $$
$$ x^2 + 34 - 3x + 20 = 0 $$
$$ x^2 - 3x + 54 = 0 $$
$$ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 54}}{2} $$
Solve for x and you have your answer

anonymous · Answer

How do you go from the third equation to the fourth one?

anonymous · Answer

That is quadratic formula.
An equation of form
$$ ax^2 + bx + c = 0 $$
has its roots as:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

anonymous · Answer

since you have the solution, you know it factors

anonymous · Answer

i.e. you already know $x=-9$ or $x=6$ so it must factor as $$x^2-3x+54=(x-9)(x+6)=0$$

phi · Answer

Apparently there is a typo.  Instead of
y=3x-20
y=x^2+34

use
y= 3x-20
y= -x^2+34  (change to -x^2 rather than +x^2)

now you should get the answers you posted.