# Using the quadratic formula, solve for x. A triangle has three angles that measure: (x+17), (3x+28), ?.The exterior angle next to ? is (x^2). What's x

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The given two angles of a triangle are x+17 and 3x+28.Therefore the remaining angle ? = 180 - sum of the given two given angles.

Therefore ? = 180 - (x+17+3x+28) = 180 - (4x+45) = 135-4x.

Therefore ? = 135-4x.

Therefore , the exterior angle of (135-4x) = 180 - (135-4x) = 45+4x.

But it is given that this angle is equal to x^2 .

So 4x+45 = x^2.

Or x^2-4x-45 = 0

x^2 -9x+5x - 45 = 0

x(x-9) +5(x-9) = 0

(x-9)(x+5) = 0.

Therefore x = 9.

The first step is to set up 2 equations.

The first is the sum of the interior angles.

(x+17)+(3x+28)+y=180

The second is the addition of 2 adjacent angles.

x^2 + y = 180

Solve the 2nd equation for y.

y=180-x^2

Then substitue into equation 1 giving

(x+17)+(3x+28)+(180-x^2)=180

by combining like terms you get

-x^2+4x+225=180

Subtract 180 from both sides

-x^2+4x+45=0

Multiple both sides by -1 in order to get x^2 term positive

x^2-4x-45=0

Now use quadratic formula

x=(-(-4) (+/-) sqrt(-4^2-4*1*-45))/(2*1)

simplified

[4 (+/-) sqrt(196)]/2

[4+/-14]/2

-10/2 = -5 18/2 = 9 answer cannot be negative so x = 9