What are the solutions to the equation 4x^2-56x+180=0 ?

The solutions to the equation 4x^2-56x+180=0 are x=9,x=5

Find the zeros of 4x^2-56x+180=0

The zeros of 4x^2-56x+180=0 are x=9,x=5

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4 x^{2} - 56 x + 180 = 0

x = 9, x = 5

Solution steps

4 x^{2} - 56 x + 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 56 ) \pm ( - 56 ) ^{2} - 4 \cdot 4 \cdot 180}{2 \cdot 4}

(- 56)^{2} - 4 \cdot 4 \cdot 180 = 16

x_{1, 2} = \frac{- ( - 56 ) \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 56 ) + 16}{2 \cdot 4}, x_{2} = \frac{- ( - 56 ) - 16}{2 \cdot 4}

x = \frac{- ( - 56 ) + 16}{2 \cdot 4} : 9

x = \frac{- ( - 56 ) - 16}{2 \cdot 4} : 5

The solutions to the quadratic equation are:

x = 9, x = 5

- 15 \leq 3 x + 6 < - 9

\frac{x}{3} + \frac{x}{2} = 15

3 x - 9 x^{2} + 10 = 0

x + 8 \leq 13

0.534 = 4.9 t^{2} + 0.442 t

What are the solutions to the equation 4x^2-56x+180=0 ?
The solutions to the equation 4x^2-56x+180=0 are x=9,x=5
Find the zeros of 4x^2-56x+180=0
The zeros of 4x^2-56x+180=0 are x=9,x=5