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

The solutions to the equation 2x^2+7=4x are x=1+isqrt(5/2),x=1-isqrt(5/2)

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

The zeros of 2x^2+7=4x are x=1+isqrt(5/2),x=1-isqrt(5/2)

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2 x^{2} + 7 = 4 x

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

Solution steps

2 x^{2} + 7 = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 7 : 210 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 10 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 10 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 10 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 10 i}{2 \cdot 2} : 1 + i \frac{5}{2}

x = \frac{- ( - 4 ) - 2 10 i}{2 \cdot 2} : 1 - i \frac{5}{2}

The solutions to the quadratic equation are:

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

\frac{2 q ^{- 9} b}{5 ( q ^{- 5} b ^{- 4} ) ^{2}}

(\frac{z}{5})^{2}

\frac{7}{12} - \frac{4}{9}

∣ x + 4 ∣ = 7.5

(\frac{a ^{- 3}}{b ^{2}})^{4}

What are the solutions to the equation 2x^2+7=4x ?
The solutions to the equation 2x^2+7=4x are x=1+isqrt(5/2),x=1-isqrt(5/2)
Find the zeros of 2x^2+7=4x
The zeros of 2x^2+7=4x are x=1+isqrt(5/2),x=1-isqrt(5/2)