What are the solutions to the equation 8r^2+23r-3=0 ?

The solutions to the equation 8r^2+23r-3=0 are r= 1/8 ,r=-3

Find the zeros of 8r^2+23r-3=0

The zeros of 8r^2+23r-3=0 are r= 1/8 ,r=-3

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8 r^{2} + 23 r - 3 = 0

r = \frac{1}{8}, r = - 3

Decimal Notation

r = 0.125, r = - 3

Solution steps

8 r^{2} + 23 r - 3 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

2 3^{2} - 4 \cdot 8 (- 3) = 25

r_{1, 2} = \frac{- 23 \pm 25}{2 \cdot 8}

Separate the solutions

r_{1} = \frac{- 23 + 25}{2 \cdot 8}, r_{2} = \frac{- 23 - 25}{2 \cdot 8}

r = \frac{- 23 + 25}{2 \cdot 8} : \frac{1}{8}

r = \frac{- 23 - 25}{2 \cdot 8} : - 3

The solutions to the quadratic equation are:

r = \frac{1}{8}, r = - 3

\frac{s}{2} + 15 \geq 11

- 6 x^{2} + 36 x - 48 = 0

t^{3} \cdot t^{4} \cdot t^{2}

- 3 x^{2} - x - 3 = 0

4 (- \frac{1}{2})

What are the solutions to the equation 8r^2+23r-3=0 ?
The solutions to the equation 8r^2+23r-3=0 are r= 1/8 ,r=-3
Find the zeros of 8r^2+23r-3=0
The zeros of 8r^2+23r-3=0 are r= 1/8 ,r=-3