What are the solutions to the equation 4x^2-36x+32=0 ?

The solutions to the equation 4x^2-36x+32=0 are x=8,x=1

Find the zeros of 4x^2-36x+32=0

The zeros of 4x^2-36x+32=0 are x=8,x=1

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4 x^{2} - 36 x + 32 = 0

x = 8, x = 1

Solution steps

4 x^{2} - 36 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 4 \cdot 32}{2 \cdot 4}

(- 36)^{2} - 4 \cdot 4 \cdot 32 = 28

x_{1, 2} = \frac{- ( - 36 ) \pm 28}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 28}{2 \cdot 4}, x_{2} = \frac{- ( - 36 ) - 28}{2 \cdot 4}

x = \frac{- ( - 36 ) + 28}{2 \cdot 4} : 8

x = \frac{- ( - 36 ) - 28}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

x = 8, x = 1

\frac{4 i}{6 + 4 i}

1 - \frac{0.01}{2}

- \frac{1}{4} - \frac{3}{8}

\frac{1}{4} - \frac{a}{2}

36 x^{3} + 9 x^{2} - 4 x - 1

What are the solutions to the equation 4x^2-36x+32=0 ?
The solutions to the equation 4x^2-36x+32=0 are x=8,x=1
Find the zeros of 4x^2-36x+32=0
The zeros of 4x^2-36x+32=0 are x=8,x=1