What are the solutions to the equation quadraticformula 2x^2-5x=-2 ?

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

Find the zeros of quadraticformula 2x^2-5x=-2

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

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2 x^{2} - 5 x = - 2

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

Solution steps

2 x^{2} - 5 x = - 2

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

x = \frac{- ( - 5 ) + 3}{2 \cdot 2} : 2

x = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

The solutions to the quadratic equation are:

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

x, 6 x + 2 + 6 x < 14

x, 7^{x} = 10

2/3.1/4

r, 2 = r - 10

x, 4 x = 6

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