What are the solutions to the equation r^2+3r-8=0 ?

The solutions to the equation r^2+3r-8=0 are r=(-3+sqrt(41))/2 ,r=(-3-sqrt(41))/2

Find the zeros of r^2+3r-8=0

The zeros of r^2+3r-8=0 are r=(-3+sqrt(41))/2 ,r=(-3-sqrt(41))/2

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r^{2} + 3 r - 8 = 0

r = \frac{- 3 + 41}{2}, r = \frac{- 3 - 41}{2}

Decimal Notation

r = 1.70156 \dots, r = - 4.70156 \dots

Solution steps

r^{2} + 3 r - 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 8) = 41

r_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 3 + 41}{2 \cdot 1}, r_{2} = \frac{- 3 - 41}{2 \cdot 1}

r = \frac{- 3 + 41}{2 \cdot 1} : \frac{- 3 + 41}{2}

r = \frac{- 3 - 41}{2 \cdot 1} : \frac{- 3 - 41}{2}

The solutions to the quadratic equation are:

r = \frac{- 3 + 41}{2}, r = \frac{- 3 - 41}{2}

- 1 + \frac{3}{2} x = 4

6 = x + 2

y^{2} - 12 = 0

4 x + 3 = \frac{5}{2} x - 7

(\frac{1}{x})^{\frac{1}{2}}

What are the solutions to the equation r^2+3r-8=0 ?
The solutions to the equation r^2+3r-8=0 are r=(-3+sqrt(41))/2 ,r=(-3-sqrt(41))/2
Find the zeros of r^2+3r-8=0
The zeros of r^2+3r-8=0 are r=(-3+sqrt(41))/2 ,r=(-3-sqrt(41))/2