What are the solutions to the equation 2x^2-4=5x ?

The solutions to the equation 2x^2-4=5x are x=(5+sqrt(57))/4 ,x=(5-sqrt(57))/4

Find the zeros of 2x^2-4=5x

The zeros of 2x^2-4=5x are x=(5+sqrt(57))/4 ,x=(5-sqrt(57))/4

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2 x^{2} - 4 = 5 x

x = \frac{5 + 57}{4}, x = \frac{5 - 57}{4}

Decimal Notation

x = 3.13745 \dots, x = - 0.63745 \dots

Solution steps

2 x^{2} - 4 = 5 x

Move

5 x

to the left side

2 x^{2} - 4 - 5 x = 0

Write in the standard form

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

2 x^{2} - 5 x - 4 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 (- 4) = 57

x_{1, 2} = \frac{- ( - 5 ) \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 57}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 57}{2 \cdot 2}

x = \frac{- ( - 5 ) + 57}{2 \cdot 2} : \frac{5 + 57}{4}

x = \frac{- ( - 5 ) - 57}{2 \cdot 2} : \frac{5 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 57}{4}, x = \frac{5 - 57}{4}

y, y^{2} - x y = 6

x, x - x^{2} - 2 x y = 0

- 4 x^{2} + 16 x + 9 = 0

4 x^{2} - 65 x + 16 = 0

x (x + 3) = 70

What are the solutions to the equation 2x^2-4=5x ?
The solutions to the equation 2x^2-4=5x are x=(5+sqrt(57))/4 ,x=(5-sqrt(57))/4
Find the zeros of 2x^2-4=5x
The zeros of 2x^2-4=5x are x=(5+sqrt(57))/4 ,x=(5-sqrt(57))/4