What are the solutions to the equation r^2-14r+48=0 ?

The solutions to the equation r^2-14r+48=0 are r=8,r=6

Find the zeros of r^2-14r+48=0

The zeros of r^2-14r+48=0 are r=8,r=6

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r^{2} - 14 r + 48 = 0

r = 8, r = 6

Solution steps

r^{2} - 14 r + 48 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 48 = 2

r_{1, 2} = \frac{- ( - 14 ) \pm 2}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 14 ) + 2}{2 \cdot 1}, r_{2} = \frac{- ( - 14 ) - 2}{2 \cdot 1}

r = \frac{- ( - 14 ) + 2}{2 \cdot 1} : 8

r = \frac{- ( - 14 ) - 2}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

r = 8, r = 6

5 x - 20

3 (w - 2) + 5 = 2 (3 w - 5)

3 - 0.3 t < 0.02 (75 - 1.8 t)

0.8 b - 6.43 = 19.9 b + 3.81 - 18.3 b

20 - 4.9 (x)^{2} = 10

What are the solutions to the equation r^2-14r+48=0 ?
The solutions to the equation r^2-14r+48=0 are r=8,r=6
Find the zeros of r^2-14r+48=0
The zeros of r^2-14r+48=0 are r=8,r=6