What are the solutions to the equation r^2+5r-3=0 ?

The solutions to the equation r^2+5r-3=0 are r=(-5+sqrt(37))/2 ,r=(-5-sqrt(37))/2

Find the zeros of r^2+5r-3=0

The zeros of r^2+5r-3=0 are r=(-5+sqrt(37))/2 ,r=(-5-sqrt(37))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

r = \frac{- 5 + 37}{2}, r = \frac{- 5 - 37}{2}

Decimal Notation

r = 0.54138 \dots, r = - 5.54138 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 3) = 37

r_{1, 2} = \frac{- 5 \pm 37}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 5 + 37}{2 \cdot 1}, r_{2} = \frac{- 5 - 37}{2 \cdot 1}

r = \frac{- 5 + 37}{2 \cdot 1} : \frac{- 5 + 37}{2}

r = \frac{- 5 - 37}{2 \cdot 1} : \frac{- 5 - 37}{2}

The solutions to the quadratic equation are:

r = \frac{- 5 + 37}{2}, r = \frac{- 5 - 37}{2}

6 x^{2} = 384

x, x^{2} - 6 x + 9 - 4 y^{2} = 0

0 = 3 k^{2} + 6000 k - 0.25 \cdot 1 0^{6}

x, x^{2} + y^{2} = cy

(x + 5)^{2} + 1 = 5

What are the solutions to the equation r^2+5r-3=0 ?
The solutions to the equation r^2+5r-3=0 are r=(-5+sqrt(37))/2 ,r=(-5-sqrt(37))/2
Find the zeros of r^2+5r-3=0
The zeros of r^2+5r-3=0 are r=(-5+sqrt(37))/2 ,r=(-5-sqrt(37))/2