What are the solutions to the equation 4x^2-5(2x-1)=0 ?

The solutions to the equation 4x^2-5(2x-1)=0 are x=(5+sqrt(5))/4 ,x=(5-sqrt(5))/4

Find the zeros of 4x^2-5(2x-1)=0

The zeros of 4x^2-5(2x-1)=0 are x=(5+sqrt(5))/4 ,x=(5-sqrt(5))/4

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

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

Decimal Notation

x = 1.80901 \dots, x = 0.69098 \dots

Solution steps

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

Expand

4 x^{2} - 5 (2 x - 1) : 4 x^{2} - 10 x + 5

4 x^{2} - 10 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

(- 10)^{2} - 4 \cdot 4 \cdot 5 = 25

x_{1, 2} = \frac{- ( - 10 ) \pm 2 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 5}{2 \cdot 4}, x_{2} = \frac{- ( - 10 ) - 2 5}{2 \cdot 4}

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

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

The solutions to the quadratic equation are:

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

x^{2} - 14 = - 5 - 5

k^{2} + 6 k - 16 = 0

x^{2} + 11 x + 30 = 0

w^{2} = 8 (w - 6)

0 = 3 x^{2} - 6 x - 1

What are the solutions to the equation 4x^2-5(2x-1)=0 ?
The solutions to the equation 4x^2-5(2x-1)=0 are x=(5+sqrt(5))/4 ,x=(5-sqrt(5))/4
Find the zeros of 4x^2-5(2x-1)=0
The zeros of 4x^2-5(2x-1)=0 are x=(5+sqrt(5))/4 ,x=(5-sqrt(5))/4