What are the solutions to the equation 4x^2-32x+28=0 ?

The solutions to the equation 4x^2-32x+28=0 are x=7,x=1

Find the zeros of 4x^2-32x+28=0

The zeros of 4x^2-32x+28=0 are x=7,x=1

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4 x^{2} - 32 x + 28 = 0

x = 7, x = 1

Solution steps

4 x^{2} - 32 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 4 \cdot 28}{2 \cdot 4}

(- 32)^{2} - 4 \cdot 4 \cdot 28 = 24

x_{1, 2} = \frac{- ( - 32 ) \pm 24}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 24}{2 \cdot 4}, x_{2} = \frac{- ( - 32 ) - 24}{2 \cdot 4}

x = \frac{- ( - 32 ) + 24}{2 \cdot 4} : 7

x = \frac{- ( - 32 ) - 24}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

x = 7, x = 1

32 - m^{5}

2 (7 - 8 x) = 32 - 10 x

2 (1 + 5 p) = - 38 + 5 p

4 x^{2} + 16 x

11 - (- 4)

What are the solutions to the equation 4x^2-32x+28=0 ?
The solutions to the equation 4x^2-32x+28=0 are x=7,x=1
Find the zeros of 4x^2-32x+28=0
The zeros of 4x^2-32x+28=0 are x=7,x=1