What are the solutions to the equation r^2+10r=-7 ?

The solutions to the equation r^2+10r=-7 are r=-5+3sqrt(2),r=-5-3sqrt(2)

Find the zeros of r^2+10r=-7

The zeros of r^2+10r=-7 are r=-5+3sqrt(2),r=-5-3sqrt(2)

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r^{2} + 10 r = - 7

r = - 5 + 32, r = - 5 - 32

Decimal Notation

r = - 0.75735 \dots, r = - 9.24264 \dots

Solution steps

r^{2} + 10 r = - 7

Move

7

to the left side

r^{2} + 10 r + 7 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot 7 = 62

r_{1, 2} = \frac{- 10 \pm 6 2}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 10 + 6 2}{2 \cdot 1}, r_{2} = \frac{- 10 - 6 2}{2 \cdot 1}

r = \frac{- 10 + 6 2}{2 \cdot 1} : - 5 + 32

r = \frac{- 10 - 6 2}{2 \cdot 1} : - 5 - 32

The solutions to the quadratic equation are:

r = - 5 + 32, r = - 5 - 32

2.32 = - 0.0007 h^{2} + 0.015 h + 3.02

t, t^{2} = 2 p + a s

x^{2} - 4 x - 10 = 2

x, 3 x^{2} - 3 x + 9 = 0

x, f (x + 3) = 4 x^{2} + 4 x - 2

What are the solutions to the equation r^2+10r=-7 ?
The solutions to the equation r^2+10r=-7 are r=-5+3sqrt(2),r=-5-3sqrt(2)
Find the zeros of r^2+10r=-7
The zeros of r^2+10r=-7 are r=-5+3sqrt(2),r=-5-3sqrt(2)