What are the solutions to the equation 2x^2-4=3x ?

The solutions to the equation 2x^2-4=3x are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4

Find the zeros of 2x^2-4=3x

The zeros of 2x^2-4=3x are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4

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2 x^{2} - 4 = 3 x

x = \frac{3 + 41}{4}, x = \frac{3 - 41}{4}

Solution steps

2 x^{2} - 4 = 3 x

Move

3 x

to the left side

2 x^{2} - 4 - 3 x = 0

Write in the standard form

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

2 x^{2} - 3 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 4) = 41

x_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 2}

x = \frac{- ( - 3 ) + 41}{2 \cdot 2} : \frac{3 + 41}{4}

x = \frac{- ( - 3 ) - 41}{2 \cdot 2} : \frac{3 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{4}, x = \frac{3 - 41}{4}

(3 x - 1) (2 x + 3) = 8

2 m^{2} + m = 1

1.5 t^{2} - 6 t + 3 = 0

t^{2} - t - 6 = 4 (t - 3)

6 x^{2} - 9 x - 15 = 0

What are the solutions to the equation 2x^2-4=3x ?
The solutions to the equation 2x^2-4=3x are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4
Find the zeros of 2x^2-4=3x
The zeros of 2x^2-4=3x are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4