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

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

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

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

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

z = \frac{- 3 + 41}{4}, z = \frac{- 3 - 41}{4}

Decimal Notation

z = 0.85078 \dots, z = - 2.35078 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = \frac{- 3 + 41}{4}, z = \frac{- 3 - 41}{4}

4 x^{2} - 12 x + 3 = 0

c = (3 (6) ln (6)) + 12

t r - 10 r + t n - 10 n

\frac{a}{3} - \frac{a}{5}

\frac{m}{- 5} \geq 2

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