What are the solutions to the equation r^2=10-3r ?

The solutions to the equation r^2=10-3r are r=2,r=-5

Find the zeros of r^2=10-3r

The zeros of r^2=10-3r are r=2,r=-5

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r^{2} = 10 - 3 r

r = 2, r = - 5

Solution steps

r^{2} = 10 - 3 r

Move

3 r

to the left side

r^{2} + 3 r = 10

Move

10

to the left side

r^{2} + 3 r - 10 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

r_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 3 + 7}{2 \cdot 1}, r_{2} = \frac{- 3 - 7}{2 \cdot 1}

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

r = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

r = 2, r = - 5

2 (x + 5)^{2} = 90

3 v^{2} = 7 v + 6

y^{2} - 7.8 y = - 12.65

x^{2} - 4 x - 10

C = (2 x - 3) (2 x - 2) - 2 x (x - 5)

What are the solutions to the equation r^2=10-3r ?
The solutions to the equation r^2=10-3r are r=2,r=-5
Find the zeros of r^2=10-3r
The zeros of r^2=10-3r are r=2,r=-5