What are the solutions to the equation 26x^2+25x-975=0 ?

The solutions to the equation 26x^2+25x-975=0 are x=(-25+5sqrt(4081))/(52),x=(-25-5sqrt(4081))/(52)

Find the zeros of 26x^2+25x-975=0

The zeros of 26x^2+25x-975=0 are x=(-25+5sqrt(4081))/(52),x=(-25-5sqrt(4081))/(52)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

26 x^{2} + 25 x - 975 = 0

x = \frac{- 25 + 5 4081}{52}, x = \frac{- 25 - 5 4081}{52}

Decimal Notation

x = 5.66179 \dots, x = - 6.62333 \dots

Solution steps

26 x^{2} + 25 x - 975 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 \cdot 26 ( - 975 )}{2 \cdot 26}

2 5^{2} - 4 \cdot 26 (- 975) = 54081

x_{1, 2} = \frac{- 25 \pm 5 4081}{2 \cdot 26}

Separate the solutions

x_{1} = \frac{- 25 + 5 4081}{2 \cdot 26}, x_{2} = \frac{- 25 - 5 4081}{2 \cdot 26}

x = \frac{- 25 + 5 4081}{2 \cdot 26} : \frac{- 25 + 5 4081}{52}

x = \frac{- 25 - 5 4081}{2 \cdot 26} : \frac{- 25 - 5 4081}{52}

The solutions to the quadratic equation are:

x = \frac{- 25 + 5 4081}{52}, x = \frac{- 25 - 5 4081}{52}

r^{2} + 15 r + 50 = 6 - r

(p + 1)^{2} = \frac{4}{9}

- 16 t^{2} + 100 = 0

\frac{785}{16} = - 16 (t)^{2} + 50 (t) + 10

32 = - 16 t^{2} + 16 t + 32

What are the solutions to the equation 26x^2+25x-975=0 ?
The solutions to the equation 26x^2+25x-975=0 are x=(-25+5sqrt(4081))/(52),x=(-25-5sqrt(4081))/(52)
Find the zeros of 26x^2+25x-975=0
The zeros of 26x^2+25x-975=0 are x=(-25+5sqrt(4081))/(52),x=(-25-5sqrt(4081))/(52)