What are the solutions to the equation 10r^2+68r-45=0 ?

The solutions to the equation 10r^2+68r-45=0 are r=(-34+sqrt(1606))/(10),r=-(34+sqrt(1606))/(10)

Find the zeros of 10r^2+68r-45=0

The zeros of 10r^2+68r-45=0 are r=(-34+sqrt(1606))/(10),r=-(34+sqrt(1606))/(10)

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10 r^{2} + 68 r - 45 = 0

r = \frac{- 34 + 1606}{10}, r = - \frac{34 + 1606}{10}

Solution steps

10 r^{2} + 68 r - 45 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 68 \pm 6 8 ^{2} - 4 \cdot 10 ( - 45 )}{2 \cdot 10}

6 8^{2} - 4 \cdot 10 (- 45) = 21606

r_{1, 2} = \frac{- 68 \pm 2 1606}{2 \cdot 10}

Separate the solutions

r_{1} = \frac{- 68 + 2 1606}{2 \cdot 10}, r_{2} = \frac{- 68 - 2 1606}{2 \cdot 10}

r = \frac{- 68 + 2 1606}{2 \cdot 10} : \frac{- 34 + 1606}{10}

r = \frac{- 68 - 2 1606}{2 \cdot 10} : - \frac{34 + 1606}{10}

The solutions to the quadratic equation are:

r = \frac{- 34 + 1606}{10}, r = - \frac{34 + 1606}{10}

(x + 1)^{2} - 2 (x - 2)^{2} = 8

2 x^{2} + 6 x + 4 = 0

2 x (x + 4) = 42

Q^{2} - 10599.74 Q + 375000 = 0

2 t^{2} + 40 t + 34 = 0

What are the solutions to the equation 10r^2+68r-45=0 ?
The solutions to the equation 10r^2+68r-45=0 are r=(-34+sqrt(1606))/(10),r=-(34+sqrt(1606))/(10)
Find the zeros of 10r^2+68r-45=0
The zeros of 10r^2+68r-45=0 are r=(-34+sqrt(1606))/(10),r=-(34+sqrt(1606))/(10)