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

The solutions to the equation 4x^2-5=1x are x= 5/4 ,x=-1

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

The zeros of 4x^2-5=1x are x= 5/4 ,x=-1

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4 x^{2} - 5 = 1 x

x = \frac{5}{4}, x = - 1

Decimal Notation

x = 1.25, x = - 1

Solution steps

4 x^{2} - 5 = 1 \cdot x

Refine

4 x^{2} - 5 = x

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 4 (- 5) = 9

x_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 4}, x_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 4}

x = \frac{- ( - 1 ) + 9}{2 \cdot 4} : \frac{5}{4}

x = \frac{- ( - 1 ) - 9}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

x = \frac{5}{4}, x = - 1

x, x^{2} - p x + 36 = 0

x, 3 x^{2} - x y^{2} = 0

(5 t - 9)^{2} - 7 (5 t - 9) + 6 = 0

x (10 - x) = 40

x, 2 x y = x^{2}

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