What are the solutions to the equation 2x^2+4=7x ?

The solutions to the equation 2x^2+4=7x are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4

Find the zeros of 2x^2+4=7x

The zeros of 2x^2+4=7x are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4

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2 x^{2} + 4 = 7 x

x = \frac{7 + 17}{4}, x = \frac{7 - 17}{4}

Decimal Notation

x = 2.78077 \dots, x = 0.71922 \dots

Solution steps

2 x^{2} + 4 = 7 x

Move

7 x

to the left side

2 x^{2} + 4 - 7 x = 0

Write in the standard form

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

2 x^{2} - 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 4 = 17

x_{1, 2} = \frac{- ( - 7 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 17}{2 \cdot 2}

x = \frac{- ( - 7 ) + 17}{2 \cdot 2} : \frac{7 + 17}{4}

x = \frac{- ( - 7 ) - 17}{2 \cdot 2} : \frac{7 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{7 + 17}{4}, x = \frac{7 - 17}{4}

\frac{m}{- 3} < 5

- 5 x = - 1

24 3^{\frac{1}{6}} \cdot 3^{\frac{1}{6}}

(70 + 22 i) - (71 + 70 i)

\frac{32}{2}

What are the solutions to the equation 2x^2+4=7x ?
The solutions to the equation 2x^2+4=7x are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4
Find the zeros of 2x^2+4=7x
The zeros of 2x^2+4=7x are x=(7+sqrt(17))/4 ,x=(7-sqrt(17))/4