What are the solutions to the equation 0=34x+60-(2x^2+200) ?

The solutions to the equation 0=34x+60-(2x^2+200) are x=7,x=10

Find the zeros of 0=34x+60-(2x^2+200)

The zeros of 0=34x+60-(2x^2+200) are x=7,x=10

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0 = 34 x + 60 - (2 x^{2} + 200)

x = 7, x = 10

Solution steps

0 = 34 x + 60 - (2 x^{2} + 200)

Expand

34 x + 60 - (2 x^{2} + 200) : - 2 x^{2} + 34 x - 140

0 = - 2 x^{2} + 34 x - 140

Switch sides

- 2 x^{2} + 34 x - 140 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 34 \pm 3 4 ^{2} - 4 ( - 2 ) ( - 140 )}{2 ( - 2 )}

3 4^{2} - 4 (- 2) (- 140) = 6

x_{1, 2} = \frac{- 34 \pm 6}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 34 + 6}{2 ( - 2 )}, x_{2} = \frac{- 34 - 6}{2 ( - 2 )}

x = \frac{- 34 + 6}{2 ( - 2 )} : 7

x = \frac{- 34 - 6}{2 ( - 2 )} : 10

The solutions to the quadratic equation are:

x = 7, x = 10

x, x^{2} + 2 y^{5} = 10 x y

\frac{14}{5} x - \frac{1}{20} x^{2} = 42.5

y - 2 = \frac{y ^{2}}{9}

10 x^{2} - 4 x + 1 = 6 x^{2}

7 m^{2} - m - 4 = 6 m^{2}

What are the solutions to the equation 0=34x+60-(2x^2+200) ?
The solutions to the equation 0=34x+60-(2x^2+200) are x=7,x=10
Find the zeros of 0=34x+60-(2x^2+200)
The zeros of 0=34x+60-(2x^2+200) are x=7,x=10