What are the solutions to the equation 1-5x+3x^2=x^2 ?

The solutions to the equation 1-5x+3x^2=x^2 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

Find the zeros of 1-5x+3x^2=x^2

The zeros of 1-5x+3x^2=x^2 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

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1 - 5 x + 3 x^{2} = x^{2}

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

Solution steps

1 - 5 x + 3 x^{2} = x^{2}

Move

x^{2}

to the left side

1 - 5 x + 2 x^{2} = 0

Write in the standard form

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

2 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

x_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

x = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

x = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

\frac{4}{5} x - \frac{1}{4} \leq \frac{7}{10} x - \frac{4}{5}

5 x^{2} - 4 x - 1 = 0

∣ 1 + 3 x ∣ - 3 \geq 2

(5 x - 4) + (9 x - 68) \geq x - 1

3 (x - 7) = - 21 + 3 x

What are the solutions to the equation 1-5x+3x^2=x^2 ?
The solutions to the equation 1-5x+3x^2=x^2 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4
Find the zeros of 1-5x+3x^2=x^2
The zeros of 1-5x+3x^2=x^2 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4