What are the solutions to the equation r^2+10r-18=61 ?

The solutions to the equation r^2+10r-18=61 are r=-5+2sqrt(26),r=-5-2sqrt(26)

Find the zeros of r^2+10r-18=61

The zeros of r^2+10r-18=61 are r=-5+2sqrt(26),r=-5-2sqrt(26)

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r^{2} + 10 r - 18 = 61

r = - 5 + 226, r = - 5 - 226

Decimal Notation

r = 5.19803 \dots, r = - 15.19803 \dots

Solution steps

r^{2} + 10 r - 18 = 61

Move

61

to the left side

r^{2} + 10 r - 79 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 79 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 79) = 426

r_{1, 2} = \frac{- 10 \pm 4 26}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 10 + 4 26}{2 \cdot 1}, r_{2} = \frac{- 10 - 4 26}{2 \cdot 1}

r = \frac{- 10 + 4 26}{2 \cdot 1} : - 5 + 226

r = \frac{- 10 - 4 26}{2 \cdot 1} : - 5 - 226

The solutions to the quadratic equation are:

r = - 5 + 226, r = - 5 - 226

h^{2} - 4 h - 32 = 0

G (x) = x^{2} + 2

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

y^{2} - 144 = 0

14 y^{2} - 3 = - 19 y

What are the solutions to the equation r^2+10r-18=61 ?
The solutions to the equation r^2+10r-18=61 are r=-5+2sqrt(26),r=-5-2sqrt(26)
Find the zeros of r^2+10r-18=61
The zeros of r^2+10r-18=61 are r=-5+2sqrt(26),r=-5-2sqrt(26)