What are the solutions to the equation -0.12x^2+10x+800=0 ?

The solutions to the equation -0.12x^2+10x+800=0 are x=-50,x= 400/3

Find the zeros of -0.12x^2+10x+800=0

The zeros of -0.12x^2+10x+800=0 are x=-50,x= 400/3

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- 0.12 x^{2} + 10 x + 800 = 0

x = - 50, x = \frac{400}{3}

Decimal Notation

x = - 50, x = 133.33333 \dots

Solution steps

- 0.12 x^{2} + 10 x + 800 = 0

Multiply both sides by

100

- 12 x^{2} + 1000 x + 80000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1000 \pm 100 0 ^{2} - 4 ( - 12 ) \cdot 80000}{2 ( - 12 )}

100 0^{2} - 4 (- 12) \cdot 80000 = 2200

x_{1, 2} = \frac{- 1000 \pm 2200}{2 ( - 12 )}

Separate the solutions

x_{1} = \frac{- 1000 + 2200}{2 ( - 12 )}, x_{2} = \frac{- 1000 - 2200}{2 ( - 12 )}

x = \frac{- 1000 + 2200}{2 ( - 12 )} : - 50

x = \frac{- 1000 - 2200}{2 ( - 12 )} : \frac{400}{3}

The solutions to the quadratic equation are:

x = - 50, x = \frac{400}{3}

0 = - 3 x^{2} - 4 x + 5

\frac{π x ^{2} - 22 x ^{2} + 720 x}{9} = 0

7 x^{2} - 70 x + 175 = 0

x^{2} + 4 x - 12 = 2

x^{2} - 3 x + 21 = 0

What are the solutions to the equation -0.12x^2+10x+800=0 ?
The solutions to the equation -0.12x^2+10x+800=0 are x=-50,x= 400/3
Find the zeros of -0.12x^2+10x+800=0
The zeros of -0.12x^2+10x+800=0 are x=-50,x= 400/3