What are the solutions to the equation 4z^2-3z-2=0 ?

The solutions to the equation 4z^2-3z-2=0 are z=(3+sqrt(41))/8 ,z=(3-sqrt(41))/8

Find the zeros of 4z^2-3z-2=0

The zeros of 4z^2-3z-2=0 are z=(3+sqrt(41))/8 ,z=(3-sqrt(41))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

z = \frac{3 + 41}{8}, z = \frac{3 - 41}{8}

Decimal Notation

z = 1.17539 \dots, z = - 0.42539 \dots

Solution steps

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

Solve with the quadratic formula

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

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

z_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 4}

Separate the solutions

z_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 4}, z_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 4}

z = \frac{- ( - 3 ) + 41}{2 \cdot 4} : \frac{3 + 41}{8}

z = \frac{- ( - 3 ) - 41}{2 \cdot 4} : \frac{3 - 41}{8}

The solutions to the quadratic equation are:

z = \frac{3 + 41}{8}, z = \frac{3 - 41}{8}

4 x - 8 = 3 x - 8

∣ 5 x - 4 ∣ = 26

(- \frac{x}{9 - x ^{2}})^{2}

x^{8} - 1

0 \leq x < 2 π

What are the solutions to the equation 4z^2-3z-2=0 ?
The solutions to the equation 4z^2-3z-2=0 are z=(3+sqrt(41))/8 ,z=(3-sqrt(41))/8
Find the zeros of 4z^2-3z-2=0
The zeros of 4z^2-3z-2=0 are z=(3+sqrt(41))/8 ,z=(3-sqrt(41))/8