What are the solutions to the equation 0=2x^2+3x-5 ?

The solutions to the equation 0=2x^2+3x-5 are x=1,x=-5/2

Find the zeros of 0=2x^2+3x-5

The zeros of 0=2x^2+3x-5 are x=1,x=-5/2

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

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

Decimal Notation

x = 1, x = - 2.5

Solution steps

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

Switch sides

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}

x^{2} - 20 x - 6 = - 3

s^{2} = 169

\frac{x ^{2}}{2} + \frac{4 x}{3} = 6

x^{2} + 9 x + 16 = 2 x + 6

5 x^{2} - 7 x + 4 = 0

What are the solutions to the equation 0=2x^2+3x-5 ?
The solutions to the equation 0=2x^2+3x-5 are x=1,x=-5/2
Find the zeros of 0=2x^2+3x-5
The zeros of 0=2x^2+3x-5 are x=1,x=-5/2