What are the solutions to the equation r^2-12r+32=0 ?

The solutions to the equation r^2-12r+32=0 are r=8,r=4

Find the zeros of r^2-12r+32=0

The zeros of r^2-12r+32=0 are r=8,r=4

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

r = 8, r = 4

Solution steps

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

x, 0 = x^{2} + x - 12

x, y (x^{2} - 4) = 1 - x^{2}

6 x^{2} - 135 = 3 x^{2}

3 x^{2} + 36 x + 1

5 x^{2} - 10 x + 19 = 0

What are the solutions to the equation r^2-12r+32=0 ?
The solutions to the equation r^2-12r+32=0 are r=8,r=4
Find the zeros of r^2-12r+32=0
The zeros of r^2-12r+32=0 are r=8,r=4