SOLUTION: can you help me solve the system algebraically y=2(x+3)^2-5 y=14x+17

Question 529656: can you help me solve the system algebraically
y=2(x+3)^2-5
y=14x+17
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
can you help me solve the system algebraically
y=2(x+3)^2-5
raise to power
2(x^2+6x+9)-5
distribute
y=2x^2+12x+18-5
combine like terms
y=2x^2+12x+13
set to zero
2x^2+12x+13=0
Solve for x using following quadratic formula:

a=2, b=12, c=13
ans:
x≈-1.42
or
x≈-4.58
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
can you help me solve the system algebraically
y=2(x+3)^2-5
y=14x+17

Since the 2 equations are equal to y, then we can say that:

(x - 2)(x + 1) = 0

x = 2 or � 1

When x = 2, then:
y = 14(2) + 17, or 28 + 17, or . Solution:

When x = - 1, then:
y = 14(- 1) + 17, or - 14 + 17, or . Solution:

You should be able to do the check to make sure that the solution pairs make the 2 equations true.

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