What are the solutions to the equation 1.2r^2+5.7r=2.3 ?

The solutions to the equation 1.2r^2+5.7r=2.3 are r=(-57+sqrt(4353))/(24),r=(-57-sqrt(4353))/(24)

Find the zeros of 1.2r^2+5.7r=2.3

The zeros of 1.2r^2+5.7r=2.3 are r=(-57+sqrt(4353))/(24),r=(-57-sqrt(4353))/(24)

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1.2 r^{2} + 5.7 r = 2.3

r = \frac{- 57 + 4353}{24}, r = \frac{- 57 - 4353}{24}

Solution steps

1.2 r^{2} + 5.7 r = 2.3

Multiply both sides by

10

12 r^{2} + 57 r = 23

Move

23

to the left side

12 r^{2} + 57 r - 23 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 57 \pm 5 7 ^{2} - 4 \cdot 12 ( - 23 )}{2 \cdot 12}

5 7^{2} - 4 \cdot 12 (- 23) = 4353

r_{1, 2} = \frac{- 57 \pm 4353}{2 \cdot 12}

Separate the solutions

r_{1} = \frac{- 57 + 4353}{2 \cdot 12}, r_{2} = \frac{- 57 - 4353}{2 \cdot 12}

r = \frac{- 57 + 4353}{2 \cdot 12} : \frac{- 57 + 4353}{24}

r = \frac{- 57 - 4353}{2 \cdot 12} : \frac{- 57 - 4353}{24}

The solutions to the quadratic equation are:

r = \frac{- 57 + 4353}{24}, r = \frac{- 57 - 4353}{24}

x^{2} - 4 x + 4 = 75

4 x^{2} + 8 x - 32 = 0

9 x^{2} - 5 x - 7 = 0

(x + 8)^{2} - 4 = - 20

x^{2} - \frac{9}{16} = 0

What are the solutions to the equation 1.2r^2+5.7r=2.3 ?
The solutions to the equation 1.2r^2+5.7r=2.3 are r=(-57+sqrt(4353))/(24),r=(-57-sqrt(4353))/(24)
Find the zeros of 1.2r^2+5.7r=2.3
The zeros of 1.2r^2+5.7r=2.3 are r=(-57+sqrt(4353))/(24),r=(-57-sqrt(4353))/(24)