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

The solutions to the equation -x^2+5x-3=0 are x=(5-sqrt(13))/2 ,x=(5+sqrt(13))/2

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

The zeros of -x^2+5x-3=0 are x=(5-sqrt(13))/2 ,x=(5+sqrt(13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{5 - 13}{2}, x = \frac{5 + 13}{2}

Decimal Notation

x = 0.69722 \dots, x = 4.30277 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 (- 1) (- 3) = 13

x_{1, 2} = \frac{- 5 \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 5 + 13}{2 ( - 1 )}, x_{2} = \frac{- 5 - 13}{2 ( - 1 )}

x = \frac{- 5 + 13}{2 ( - 1 )} : \frac{5 - 13}{2}

x = \frac{- 5 - 13}{2 ( - 1 )} : \frac{5 + 13}{2}

The solutions to the quadratic equation are:

x = \frac{5 - 13}{2}, x = \frac{5 + 13}{2}

x + 10 x^{2} - 12 x = x^{2} + 2 x

x^{2} - \frac{9}{2} x - \frac{24}{5} = 0

x, x^{2} + y^{2} - 4 x + 12 y + 40 = 0

r, (x - h)^{2} + (y - h)^{2} = r^{2}

4 x^{2} - 18 x + 18 = 0

What are the solutions to the equation -x^2+5x-3=0 ?
The solutions to the equation -x^2+5x-3=0 are x=(5-sqrt(13))/2 ,x=(5+sqrt(13))/2
Find the zeros of -x^2+5x-3=0
The zeros of -x^2+5x-3=0 are x=(5-sqrt(13))/2 ,x=(5+sqrt(13))/2