What are the solutions to the equation -z^2-3z-13=-2z^2 ?

The solutions to the equation -z^2-3z-13=-2z^2 are z=(3+sqrt(61))/2 ,z=(3-sqrt(61))/2

Find the zeros of -z^2-3z-13=-2z^2

The zeros of -z^2-3z-13=-2z^2 are z=(3+sqrt(61))/2 ,z=(3-sqrt(61))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- z^{2} - 3 z - 13 = - 2 z^{2}

z = \frac{3 + 61}{2}, z = \frac{3 - 61}{2}

Decimal Notation

z = 5.40512 \dots, z = - 2.40512 \dots

Solution steps

- z^{2} - 3 z - 13 = - 2 z^{2}

Move

2 z^{2}

to the left side

z^{2} - 3 z - 13 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 13) = 61

z_{1, 2} = \frac{- ( - 3 ) \pm 61}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 3 ) + 61}{2 \cdot 1}, z_{2} = \frac{- ( - 3 ) - 61}{2 \cdot 1}

z = \frac{- ( - 3 ) + 61}{2 \cdot 1} : \frac{3 + 61}{2}

z = \frac{- ( - 3 ) - 61}{2 \cdot 1} : \frac{3 - 61}{2}

The solutions to the quadratic equation are:

z = \frac{3 + 61}{2}, z = \frac{3 - 61}{2}

10 = x^{2} - 9 x

5 x^{2} - 90 x + 310 = 0

x^{2} + 12 x - 85 = 0

- 2 x^{2} + 20 x - 22 = 0

5 x^{2} + 10 x = 75

What are the solutions to the equation -z^2-3z-13=-2z^2 ?
The solutions to the equation -z^2-3z-13=-2z^2 are z=(3+sqrt(61))/2 ,z=(3-sqrt(61))/2
Find the zeros of -z^2-3z-13=-2z^2
The zeros of -z^2-3z-13=-2z^2 are z=(3+sqrt(61))/2 ,z=(3-sqrt(61))/2