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

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

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

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

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2 x^{2} + 3 x = 5

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

Solution steps

2 x^{2} + 3 x = 5

Move

5

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}

\frac{4}{81}

2 32 a^{3} b^{5} \cdot 8 a^{7} b^{2}

- 7 \cdot 8

x - 7 < 3 x + 5 \leq 2 x + 11

(5 - 2) (5 + 2)

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