What are the solutions to the equation 2x^2+5x-4=-1 ?

The solutions to the equation 2x^2+5x-4=-1 are x= 1/2 ,x=-3

Find the zeros of 2x^2+5x-4=-1

The zeros of 2x^2+5x-4=-1 are x= 1/2 ,x=-3

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2 x^{2} + 5 x - 4 = - 1

x = \frac{1}{2}, x = - 3

Decimal Notation

x = 0.5, x = - 3

Solution steps

2 x^{2} + 5 x - 4 = - 1

Move

1

to the left side

2 x^{2} + 5 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 3) = 7

x_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 7}{2 \cdot 2}, x_{2} = \frac{- 5 - 7}{2 \cdot 2}

x = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

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

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 3

- 7 w^{2} + 30 w - 8

- \frac{3 ( - 5 ) ( - 3 ) ^{2}}{2 ( - 3 ) ^{3} - ( - 5 )}

3 b^{2} - 27 b = 0

x^{4} x^{2}

What are the solutions to the equation 2x^2+5x-4=-1 ?
The solutions to the equation 2x^2+5x-4=-1 are x= 1/2 ,x=-3
Find the zeros of 2x^2+5x-4=-1
The zeros of 2x^2+5x-4=-1 are x= 1/2 ,x=-3