What are the solutions to the equation 25r^2+175r-168=0 ?

The solutions to the equation 25r^2+175r-168=0 are r=(-35+sqrt(1897))/(10),r=-(35+sqrt(1897))/(10)

Find the zeros of 25r^2+175r-168=0

The zeros of 25r^2+175r-168=0 are r=(-35+sqrt(1897))/(10),r=-(35+sqrt(1897))/(10)

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25 r^{2} + 175 r - 168 = 0

r = \frac{- 35 + 1897}{10}, r = - \frac{35 + 1897}{10}

Decimal Notation

r = 0.85545 \dots, r = - 7.85545 \dots

Solution steps

25 r^{2} + 175 r - 168 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 175 \pm 17 5 ^{2} - 4 \cdot 25 ( - 168 )}{2 \cdot 25}

17 5^{2} - 4 \cdot 25 (- 168) = 51897

r_{1, 2} = \frac{- 175 \pm 5 1897}{2 \cdot 25}

Separate the solutions

r_{1} = \frac{- 175 + 5 1897}{2 \cdot 25}, r_{2} = \frac{- 175 - 5 1897}{2 \cdot 25}

r = \frac{- 175 + 5 1897}{2 \cdot 25} : \frac{- 35 + 1897}{10}

r = \frac{- 175 - 5 1897}{2 \cdot 25} : - \frac{35 + 1897}{10}

The solutions to the quadratic equation are:

r = \frac{- 35 + 1897}{10}, r = - \frac{35 + 1897}{10}

10 r^{2} - (10 + 2 (10 r - 10)) = 2.5

a (x + b)^{2} = c

b \cdot b = 288 - 98 + 200 - 162

g, x lo g_{2} (x + 6) = 3 - lo g_{2} (x + 4)

u, v^{2} = u^{2} + 2 a^{3}

What are the solutions to the equation 25r^2+175r-168=0 ?
The solutions to the equation 25r^2+175r-168=0 are r=(-35+sqrt(1897))/(10),r=-(35+sqrt(1897))/(10)
Find the zeros of 25r^2+175r-168=0
The zeros of 25r^2+175r-168=0 are r=(-35+sqrt(1897))/(10),r=-(35+sqrt(1897))/(10)