What are the solutions to the equation 4x^2-2x=5 ?

The solutions to the equation 4x^2-2x=5 are x=(1+sqrt(21))/4 ,x=(1-sqrt(21))/4

Find the zeros of 4x^2-2x=5

The zeros of 4x^2-2x=5 are x=(1+sqrt(21))/4 ,x=(1-sqrt(21))/4

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4 x^{2} - 2 x = 5

x = \frac{1 + 21}{4}, x = \frac{1 - 21}{4}

Decimal Notation

x = 1.39564 \dots, x = - 0.89564 \dots

Solution steps

4 x^{2} - 2 x = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 4 (- 5) = 221

x_{1, 2} = \frac{- ( - 2 ) \pm 2 21}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 21}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 2 21}{2 \cdot 4}

x = \frac{- ( - 2 ) + 2 21}{2 \cdot 4} : \frac{1 + 21}{4}

x = \frac{- ( - 2 ) - 2 21}{2 \cdot 4} : \frac{1 - 21}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 21}{4}, x = \frac{1 - 21}{4}

w^{2} - 14 w + 2 = - 4 w

7 x^{2} + 8 x + 3 = 0

3 x^{2} - 9 x - 84 = 0

8 x^{2} = - 128

y, x^{2} + y^{2} - 5 = 2 x y + y^{2} - 5

What are the solutions to the equation 4x^2-2x=5 ?
The solutions to the equation 4x^2-2x=5 are x=(1+sqrt(21))/4 ,x=(1-sqrt(21))/4
Find the zeros of 4x^2-2x=5
The zeros of 4x^2-2x=5 are x=(1+sqrt(21))/4 ,x=(1-sqrt(21))/4