What are the solutions to the equation x^2-5x+25/4 =4 ?

The solutions to the equation x^2-5x+25/4 =4 are x= 9/2 ,x= 1/2

Find the zeros of x^2-5x+25/4 =4

The zeros of x^2-5x+25/4 =4 are x= 9/2 ,x= 1/2

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x^{2} - 5 x + \frac{25}{4} = 4

x = \frac{9}{2}, x = \frac{1}{2}

Decimal Notation

x = 4.5, x = 0.5

Solution steps

x^{2} - 5 x + \frac{25}{4} = 4

Multiply both sides by

4

4 x^{2} - 20 x + 25 = 16

Move

16

to the left side

4 x^{2} - 20 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 9 = 16

x_{1, 2} = \frac{- ( - 20 ) \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 16}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 16}{2 \cdot 4}

x = \frac{- ( - 20 ) + 16}{2 \cdot 4} : \frac{9}{2}

x = \frac{- ( - 20 ) - 16}{2 \cdot 4} : \frac{1}{2}

The solutions to the quadratic equation are:

x = \frac{9}{2}, x = \frac{1}{2}

2 y^{2} - 14 = 63

a^{2} - 3 a + 3 = 3

2 v^{2} + v + 1 = 0

3 c^{2} - 13 c + 4 = - 4

y, x^{3} y^{2} = x^{2} + y^{2}

What are the solutions to the equation x^2-5x+25/4 =4 ?
The solutions to the equation x^2-5x+25/4 =4 are x= 9/2 ,x= 1/2
Find the zeros of x^2-5x+25/4 =4
The zeros of x^2-5x+25/4 =4 are x= 9/2 ,x= 1/2