What are the solutions to the equation -16x^2+240x+180=0 ?

The solutions to the equation -16x^2+240x+180=0 are x=-(3(sqrt(30)-5))/2 ,x=(3(5+sqrt(30)))/2

Find the zeros of -16x^2+240x+180=0

The zeros of -16x^2+240x+180=0 are x=-(3(sqrt(30)-5))/2 ,x=(3(5+sqrt(30)))/2

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- 16 x^{2} + 240 x + 180 = 0

x = - \frac{3 ( 30 - 5 )}{2}, x = \frac{3 ( 5 + 30 )}{2}

Decimal Notation

x = - 0.71583 \dots, x = 15.71583 \dots

Solution steps

- 16 x^{2} + 240 x + 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 240 \pm 24 0 ^{2} - 4 ( - 16 ) \cdot 180}{2 ( - 16 )}

24 0^{2} - 4 (- 16) \cdot 180 = 4830

x_{1, 2} = \frac{- 240 \pm 48 30}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 240 + 48 30}{2 ( - 16 )}, x_{2} = \frac{- 240 - 48 30}{2 ( - 16 )}

x = \frac{- 240 + 48 30}{2 ( - 16 )} : - \frac{3 ( 30 - 5 )}{2}

x = \frac{- 240 - 48 30}{2 ( - 16 )} : \frac{3 ( 5 + 30 )}{2}

The solutions to the quadratic equation are:

x = - \frac{3 ( 30 - 5 )}{2}, x = \frac{3 ( 5 + 30 )}{2}

9 x^{2} - 30 x - 25 = 0

9 x^{2} - 8 x = 0

4 x^{2} + 8 x + 80 = 0

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

14 = x^{2} + 5 x

What are the solutions to the equation -16x^2+240x+180=0 ?
The solutions to the equation -16x^2+240x+180=0 are x=-(3(sqrt(30)-5))/2 ,x=(3(5+sqrt(30)))/2
Find the zeros of -16x^2+240x+180=0
The zeros of -16x^2+240x+180=0 are x=-(3(sqrt(30)-5))/2 ,x=(3(5+sqrt(30)))/2