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

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

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

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

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

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

Decimal Notation

x = 0.85078 \dots, x = - 2.35078 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 (x + 5) = 6 x + 4

\frac{2 x - 7}{3} = 5

\frac{6 x ^{3} y}{12 x ^{2} y}

7 \cdot \frac{1}{7}

87.1 = \frac{[ ( 1.96 + x ) 10 ] ^{2}}{( 15 - 12 ) ^{2}}

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