What are the solutions to the equation 28=5r+3r^2 ?

The solutions to the equation 28=5r+3r^2 are r= 7/3 ,r=-4

Find the zeros of 28=5r+3r^2

The zeros of 28=5r+3r^2 are r= 7/3 ,r=-4

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28 = 5 r + 3 r^{2}

r = \frac{7}{3}, r = - 4

Decimal Notation

r = 2.33333 \dots, r = - 4

Solution steps

28 = 5 r + 3 r^{2}

Switch sides

5 r + 3 r^{2} = 28

Move

28

to the left side

5 r + 3 r^{2} - 28 = 0

Write in the standard form

a x^{2} + b x + c = 0

3 r^{2} + 5 r - 28 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 ( - 28 )}{2 \cdot 3}

5^{2} - 4 \cdot 3 (- 28) = 19

r_{1, 2} = \frac{- 5 \pm 19}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- 5 + 19}{2 \cdot 3}, r_{2} = \frac{- 5 - 19}{2 \cdot 3}

r = \frac{- 5 + 19}{2 \cdot 3} : \frac{7}{3}

r = \frac{- 5 - 19}{2 \cdot 3} : - 4

The solutions to the quadratic equation are:

r = \frac{7}{3}, r = - 4

x, 9 x^{2} + 36 y^{2} = 0

2 x^{2} + 4 x + 38 = - 6

6 x^{2} + 72 x + 600 = 0

x^{2} + 12 x - 44 = 64

3 x (x - 4) = 6

What are the solutions to the equation 28=5r+3r^2 ?
The solutions to the equation 28=5r+3r^2 are r= 7/3 ,r=-4
Find the zeros of 28=5r+3r^2
The zeros of 28=5r+3r^2 are r= 7/3 ,r=-4