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

The solutions to the equation x^2+3=4x are x=3,x=1

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

The zeros of x^2+3=4x are x=3,x=1

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x^{2} + 3 = 4 x

x = 3, x = 1

Solution steps

x^{2} + 3 = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

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

The solutions to the quadratic equation are:

x = 3, x = 1

(2 x + 3 i) (2 x - 3 i)

3 x^{2} - 12 x + 13 = 0

12 = 2 + 27 t - 16 t^{2}

5^{2} + b^{2} = 6^{2}

What are the solutions to the equation x^2+3=4x ?
The solutions to the equation x^2+3=4x are x=3,x=1
Find the zeros of x^2+3=4x
The zeros of x^2+3=4x are x=3,x=1