What are the solutions to the equation r^2-51r+270=0 ?

The solutions to the equation r^2-51r+270=0 are r=45,r=6

Find the zeros of r^2-51r+270=0

The zeros of r^2-51r+270=0 are r=45,r=6

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r^{2} - 51 r + 270 = 0

r = 45, r = 6

Solution steps

r^{2} - 51 r + 270 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 51 ) \pm ( - 51 ) ^{2} - 4 \cdot 1 \cdot 270}{2 \cdot 1}

(- 51)^{2} - 4 \cdot 1 \cdot 270 = 39

r_{1, 2} = \frac{- ( - 51 ) \pm 39}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 51 ) + 39}{2 \cdot 1}, r_{2} = \frac{- ( - 51 ) - 39}{2 \cdot 1}

r = \frac{- ( - 51 ) + 39}{2 \cdot 1} : 45

r = \frac{- ( - 51 ) - 39}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

r = 45, r = 6

1 1^{2} + 7^{2} = c^{2}

0 = 0.5 x^{2} + 15 x + 5000

x (x + 12) = - 35

x^{2} - 16 x + 28 = 0

x^{2} + 0.07 = 2.4 x

What are the solutions to the equation r^2-51r+270=0 ?
The solutions to the equation r^2-51r+270=0 are r=45,r=6
Find the zeros of r^2-51r+270=0
The zeros of r^2-51r+270=0 are r=45,r=6