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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 41}{2 \cdot 4}, x_{2} = \frac{- 3 - 41}{2 \cdot 4}

x = \frac{- 3 + 41}{2 \cdot 4} : \frac{- 3 + 41}{8}

x = \frac{- 3 - 41}{2 \cdot 4} : \frac{- 3 - 41}{8}

The solutions to the quadratic equation are:

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

4 (3 t - 9) = 3 (4 t - 12)

(2 x - 3)^{2} - 72 = 0

x^{2} + 16 x y + 63 y^{2}

q^{2} = - 81

4 x^{2} + 35 x + 24 = 0

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