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

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

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

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

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

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

Decimal Notation

q = 0.85078 \dots, q = - 2.35078 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x^{2} y^{3} (2 x^{3})

\frac{2}{1 + 3}

\frac{a}{10} - \frac{a}{6} = 1

∣ x - 6 ∣ \leq 3

k^{3} - 64

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