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

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

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

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

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5 - 3 x = 2 x^{2}

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

Decimal

x = 1, x = - 2.5

Solution steps

5 - 3 x = 2 x^{2}

Switch sides

2 x^{2} = 5 - 3 x

Move

3 x

to the left side

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}

x^{2} + (d - 24) x + 8 d = 0

x^{2} + x^{2} = 25

u^{2} = - 11 u

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

2 x^{2} + x - 300 = 0

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