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

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

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

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

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

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

Decimal Notation

x = 1, x = - 2.5

Solution steps

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

Move

3 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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}

5 x^{2} - 10 x + 3

(7 x + 8) (2 x - 11) = 0

x^{2} - 5 = - 4

A, B = A^{2} - 3 A + 21

(2 x + 3)^{2} + (x + 4)^{2} = 10

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