What are the solutions to the equation (36-2r)^2=36+2r ?

The solutions to the equation (36-2r)^2=36+2r are r= 45/2 ,r=14

Find the zeros of (36-2r)^2=36+2r

The zeros of (36-2r)^2=36+2r are r= 45/2 ,r=14

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(36 - 2 r)^{2} = 36 + 2 r

r = \frac{45}{2}, r = 14

Decimal Notation

r = 22.5, r = 14

Solution steps

(36 - 2 r)^{2} = 36 + 2 r

Expand

(36 - 2 r)^{2} : 1296 - 144 r + 4 r^{2}

1296 - 144 r + 4 r^{2} = 36 + 2 r

Move

2 r

to the left side

4 r^{2} - 146 r + 1296 = 36

Move

36

to the left side

4 r^{2} - 146 r + 1260 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 146 ) \pm ( - 146 ) ^{2} - 4 \cdot 4 \cdot 1260}{2 \cdot 4}

(- 146)^{2} - 4 \cdot 4 \cdot 1260 = 34

r_{1, 2} = \frac{- ( - 146 ) \pm 34}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- ( - 146 ) + 34}{2 \cdot 4}, r_{2} = \frac{- ( - 146 ) - 34}{2 \cdot 4}

r = \frac{- ( - 146 ) + 34}{2 \cdot 4} : \frac{45}{2}

r = \frac{- ( - 146 ) - 34}{2 \cdot 4} : 14

The solutions to the quadratic equation are:

r = \frac{45}{2}, r = 14

(x + 5)^{2} = x + 7

8 x^{2} + 3 - 6 x = 5 x^{2}

3 (y - 4)^{2} = 192

- 3 x + 3 = x^{2} - x

5 x^{2} + 48 x + 27 = (5 x + 3)

What are the solutions to the equation (36-2r)^2=36+2r ?
The solutions to the equation (36-2r)^2=36+2r are r= 45/2 ,r=14
Find the zeros of (36-2r)^2=36+2r
The zeros of (36-2r)^2=36+2r are r= 45/2 ,r=14