What are the solutions to the equation r^2+32=12r ?

The solutions to the equation r^2+32=12r are r=8,r=4

Find the zeros of r^2+32=12r

The zeros of r^2+32=12r are r=8,r=4

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r^{2} + 32 = 12 r

r = 8, r = 4

Solution steps

r^{2} + 32 = 12 r

Move

12 r

to the left side

r^{2} + 32 - 12 r = 0

Write in the standard form

a x^{2} + b x + c = 0

r^{2} - 12 r + 32 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

r_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, r_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

r = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

r = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

r = 8, r = 4

5200 + 25 x + \frac{1}{2} x^{2} = (195 - \frac{1}{2} x) x

5 x^{2} - 3 x = 8

(x + 4)^{2} - 1 = 8

2 x^{2} = - 4 x + 5

2 x^{2} = - 4 x + 3

What are the solutions to the equation r^2+32=12r ?
The solutions to the equation r^2+32=12r are r=8,r=4
Find the zeros of r^2+32=12r
The zeros of r^2+32=12r are r=8,r=4