What are the solutions to the equation 4x^2-36x+80=0 ?

The solutions to the equation 4x^2-36x+80=0 are x=5,x=4

Find the zeros of 4x^2-36x+80=0

The zeros of 4x^2-36x+80=0 are x=5,x=4

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4 x^{2} - 36 x + 80 = 0

x = 5, x = 4

Solution steps

4 x^{2} - 36 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 4 \cdot 80}{2 \cdot 4}

(- 36)^{2} - 4 \cdot 4 \cdot 80 = 4

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

Separate the solutions

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

x = \frac{- ( - 36 ) + 4}{2 \cdot 4} : 5

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

The solutions to the quadratic equation are:

x = 5, x = 4

172

3 y - 1 < - 10 and 2 y + 2 \leq - 10

(2 x + 5) (x - 4) = - 5

x^{2} + x = 30

\frac{5}{2} x + \frac{1}{2} x = 5 + \frac{7}{2} x

What are the solutions to the equation 4x^2-36x+80=0 ?
The solutions to the equation 4x^2-36x+80=0 are x=5,x=4
Find the zeros of 4x^2-36x+80=0
The zeros of 4x^2-36x+80=0 are x=5,x=4