What are the solutions to the equation 16x^2-120x+180=0 ?

The solutions to the equation 16x^2-120x+180=0 are x=(3(5+sqrt(5)))/4 ,x=(3(5-sqrt(5)))/4

Find the zeros of 16x^2-120x+180=0

The zeros of 16x^2-120x+180=0 are x=(3(5+sqrt(5)))/4 ,x=(3(5-sqrt(5)))/4

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16 x^{2} - 120 x + 180 = 0

x = \frac{3 ( 5 + 5 )}{4}, x = \frac{3 ( 5 - 5 )}{4}

Decimal Notation

x = 5.42705 \dots, x = 2.07294 \dots

Solution steps

16 x^{2} - 120 x + 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 120 ) \pm ( - 120 ) ^{2} - 4 \cdot 16 \cdot 180}{2 \cdot 16}

(- 120)^{2} - 4 \cdot 16 \cdot 180 = 245

x_{1, 2} = \frac{- ( - 120 ) \pm 24 5}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 120 ) + 24 5}{2 \cdot 16}, x_{2} = \frac{- ( - 120 ) - 24 5}{2 \cdot 16}

x = \frac{- ( - 120 ) + 24 5}{2 \cdot 16} : \frac{3 ( 5 + 5 )}{4}

x = \frac{- ( - 120 ) - 24 5}{2 \cdot 16} : \frac{3 ( 5 - 5 )}{4}

The solutions to the quadratic equation are:

x = \frac{3 ( 5 + 5 )}{4}, x = \frac{3 ( 5 - 5 )}{4}

x^{2} - 15 x + 26 = - 3 x - 6

8 x^{2} + 24 x - 20 = 0

2 x^{2} - 5 x - 7 = 0

(- x - 1)^{2} = 0

x, f (x yz) = x^{2} + y^{2} + z^{2}

What are the solutions to the equation 16x^2-120x+180=0 ?
The solutions to the equation 16x^2-120x+180=0 are x=(3(5+sqrt(5)))/4 ,x=(3(5-sqrt(5)))/4
Find the zeros of 16x^2-120x+180=0
The zeros of 16x^2-120x+180=0 are x=(3(5+sqrt(5)))/4 ,x=(3(5-sqrt(5)))/4