What are the solutions to the equation 3r^2-4050r+1362500=0 ?

The solutions to the equation 3r^2-4050r+1362500=0 are r=(25(81+sqrt(21)))/3 ,r=(25(81-sqrt(21)))/3

Find the zeros of 3r^2-4050r+1362500=0

The zeros of 3r^2-4050r+1362500=0 are r=(25(81+sqrt(21)))/3 ,r=(25(81-sqrt(21)))/3

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3 r^{2} - 4050 r + 1362500 = 0

r = \frac{25 ( 81 + 21 )}{3}, r = \frac{25 ( 81 - 21 )}{3}

Decimal Notation

r = 713.18813 \dots, r = 636.81186 \dots

Solution steps

3 r^{2} - 4050 r + 1362500 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4050 ) \pm ( - 4050 ) ^{2} - 4 \cdot 3 \cdot 1362500}{2 \cdot 3}

(- 4050)^{2} - 4 \cdot 3 \cdot 1362500 = 5021

r_{1, 2} = \frac{- ( - 4050 ) \pm 50 21}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- ( - 4050 ) + 50 21}{2 \cdot 3}, r_{2} = \frac{- ( - 4050 ) - 50 21}{2 \cdot 3}

r = \frac{- ( - 4050 ) + 50 21}{2 \cdot 3} : \frac{25 ( 81 + 21 )}{3}

r = \frac{- ( - 4050 ) - 50 21}{2 \cdot 3} : \frac{25 ( 81 - 21 )}{3}

The solutions to the quadratic equation are:

r = \frac{25 ( 81 + 21 )}{3}, r = \frac{25 ( 81 - 21 )}{3}

(3 x - 4)^{2} + 17 x = 35

- 6 x^{2} = - 14 x

x^{2} - 28 x - 60 = 0

n, d = \frac{n ( n - 3 )}{2}

2 b^{2} + 15 = 16 b

What are the solutions to the equation 3r^2-4050r+1362500=0 ?
The solutions to the equation 3r^2-4050r+1362500=0 are r=(25(81+sqrt(21)))/3 ,r=(25(81-sqrt(21)))/3
Find the zeros of 3r^2-4050r+1362500=0
The zeros of 3r^2-4050r+1362500=0 are r=(25(81+sqrt(21)))/3 ,r=(25(81-sqrt(21)))/3