What are the solutions to the equation 24^2+(13+r)^2=(17-3r)^2 ?

The solutions to the equation 24^2+(13+r)^2=(17-3r)^2 are r=-3,r=19

Find the zeros of 24^2+(13+r)^2=(17-3r)^2

The zeros of 24^2+(13+r)^2=(17-3r)^2 are r=-3,r=19

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2 4^{2} + (13 + r)^{2} = (17 - 3 r)^{2}

r = - 3, r = 19

Solution steps

2 4^{2} + (13 + r)^{2} = (17 - 3 r)^{2}

Expand

2 4^{2} + (13 + r)^{2} : r^{2} + 26 r + 745

Expand

(17 - 3 r)^{2} : 289 - 102 r + 9 r^{2}

r^{2} + 26 r + 745 = 289 - 102 r + 9 r^{2}

Move

9 r^{2}

to the left side

- 8 r^{2} + 26 r + 745 = 289 - 102 r

Move

102 r

to the left side

- 8 r^{2} + 128 r + 745 = 289

Move

289

to the left side

- 8 r^{2} + 128 r + 456 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 8 ) \cdot 456}{2 ( - 8 )}

12 8^{2} - 4 (- 8) \cdot 456 = 176

r_{1, 2} = \frac{- 128 \pm 176}{2 ( - 8 )}

Separate the solutions

r_{1} = \frac{- 128 + 176}{2 ( - 8 )}, r_{2} = \frac{- 128 - 176}{2 ( - 8 )}

r = \frac{- 128 + 176}{2 ( - 8 )} : - 3

r = \frac{- 128 - 176}{2 ( - 8 )} : 19

The solutions to the quadratic equation are:

r = - 3, r = 19

24 m^{2} + 18 m - 15

2 (- 2) + 3

25 x - 5 \leq 10 - 20 x

x^{2} - 42 = x

(\frac{2 b}{a ^{2}})^{5}

What are the solutions to the equation 24^2+(13+r)^2=(17-3r)^2 ?
The solutions to the equation 24^2+(13+r)^2=(17-3r)^2 are r=-3,r=19
Find the zeros of 24^2+(13+r)^2=(17-3r)^2
The zeros of 24^2+(13+r)^2=(17-3r)^2 are r=-3,r=19