What are the solutions to the equation 2x^2+3x-1=4 ?

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

Find the zeros of 2x^2+3x-1=4

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

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2 x^{2} + 3 x - 1 = 4

x = 1, x = - \frac{5}{2}

Decimal Notation

x = 1, x = - 2.5

Solution steps

2 x^{2} + 3 x - 1 = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 7}{2 \cdot 2}, x_{2} = \frac{- 3 - 7}{2 \cdot 2}

x = \frac{- 3 + 7}{2 \cdot 2} : 1

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

The solutions to the quadratic equation are:

x = 1, x = - \frac{5}{2}

8 x^{2} + 30 x + 13

(x - 5)^{2}

64 u^{6} - 27 w^{6}

- 6 ∣ 5 - 2 y ∣ = - 9

5 (- 2 n + 4) + 2 (n + 3)

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