What are the solutions to the equation 15r^2+7r-4=0 ?

The solutions to the equation 15r^2+7r-4=0 are r= 1/3 ,r=-4/5

Find the zeros of 15r^2+7r-4=0

The zeros of 15r^2+7r-4=0 are r= 1/3 ,r=-4/5

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15 r^{2} + 7 r - 4 = 0

r = \frac{1}{3}, r = - \frac{4}{5}

Decimal Notation

r = 0.33333 \dots, r = - 0.8

Solution steps

15 r^{2} + 7 r - 4 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 15 ( - 4 )}{2 \cdot 15}

7^{2} - 4 \cdot 15 (- 4) = 17

r_{1, 2} = \frac{- 7 \pm 17}{2 \cdot 15}

Separate the solutions

r_{1} = \frac{- 7 + 17}{2 \cdot 15}, r_{2} = \frac{- 7 - 17}{2 \cdot 15}

r = \frac{- 7 + 17}{2 \cdot 15} : \frac{1}{3}

r = \frac{- 7 - 17}{2 \cdot 15} : - \frac{4}{5}

The solutions to the quadratic equation are:

r = \frac{1}{3}, r = - \frac{4}{5}

20 - 4

12 x + 6

3 (4 x + 6) = 9 x + 12

4 u^{2} + 28 = 0

9 x^{2} + 9 x - 4 = 0

What are the solutions to the equation 15r^2+7r-4=0 ?
The solutions to the equation 15r^2+7r-4=0 are r= 1/3 ,r=-4/5
Find the zeros of 15r^2+7r-4=0
The zeros of 15r^2+7r-4=0 are r= 1/3 ,r=-4/5