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

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

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

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

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2 x^{2} - 5 x = 4

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

Solution steps

2 x^{2} - 5 x = 4

Move

4

to the left side

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}

9^{5}

x^{- 5} \cdot x^{3}

736

2 X^{2} + 4 X - 6 = 0

\frac{3 x - 2}{5} \geq \frac{2 x - 4}{6}

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