What are the solutions to the equation 4x^2-20x+25=7 ?

The solutions to the equation 4x^2-20x+25=7 are x=(5+sqrt(7))/2 ,x=(5-sqrt(7))/2

Find the zeros of 4x^2-20x+25=7

The zeros of 4x^2-20x+25=7 are x=(5+sqrt(7))/2 ,x=(5-sqrt(7))/2

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4 x^{2} - 20 x + 25 = 7

x = \frac{5 + 7}{2}, x = \frac{5 - 7}{2}

Solution steps

4 x^{2} - 20 x + 25 = 7

Move

7

to the left side

4 x^{2} - 20 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 18}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 18 = 47

x_{1, 2} = \frac{- ( - 20 ) \pm 4 7}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 4 7}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 4 7}{2 \cdot 4}

x = \frac{- ( - 20 ) + 4 7}{2 \cdot 4} : \frac{5 + 7}{2}

x = \frac{- ( - 20 ) - 4 7}{2 \cdot 4} : \frac{5 - 7}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 7}{2}, x = \frac{5 - 7}{2}

x^{2} + 16 x = - 139

- 2 x + 6 = 24

4 x^{1} < - 11

x^{2} + x - 31 = 0

- 3 + 2 x > 9

What are the solutions to the equation 4x^2-20x+25=7 ?
The solutions to the equation 4x^2-20x+25=7 are x=(5+sqrt(7))/2 ,x=(5-sqrt(7))/2
Find the zeros of 4x^2-20x+25=7
The zeros of 4x^2-20x+25=7 are x=(5+sqrt(7))/2 ,x=(5-sqrt(7))/2