What are the solutions to the equation (117-x)(333-x)-81^2=0 ?

The solutions to the equation (117-x)(333-x)-81^2=0 are x=360,x=90

Find the zeros of (117-x)(333-x)-81^2=0

The zeros of (117-x)(333-x)-81^2=0 are x=360,x=90

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(117 - x) (333 - x) - 8 1^{2} = 0

x = 360, x = 90

Solution steps

(117 - x) (333 - x) - 8 1^{2} = 0

Expand

(117 - x) (333 - x) - 8 1^{2} : x^{2} - 450 x + 32400

x^{2} - 450 x + 32400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 450 ) \pm ( - 450 ) ^{2} - 4 \cdot 1 \cdot 32400}{2 \cdot 1}

(- 450)^{2} - 4 \cdot 1 \cdot 32400 = 270

x_{1, 2} = \frac{- ( - 450 ) \pm 270}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 450 ) + 270}{2 \cdot 1}, x_{2} = \frac{- ( - 450 ) - 270}{2 \cdot 1}

x = \frac{- ( - 450 ) + 270}{2 \cdot 1} : 360

x = \frac{- ( - 450 ) - 270}{2 \cdot 1} : 90

The solutions to the quadratic equation are:

x = 360, x = 90

x, (2 x - 1)^{2} = 9 (y^{2} + z^{2})

b, z = - \frac{b ^{2} x}{y}

- 8 x^{2} + 10 x = 0

x^{2} - 10 x - 67 = 8

x, 4 x^{2} + y^{2} + z^{2} - 8 x = 0

What are the solutions to the equation (117-x)(333-x)-81^2=0 ?
The solutions to the equation (117-x)(333-x)-81^2=0 are x=360,x=90
Find the zeros of (117-x)(333-x)-81^2=0
The zeros of (117-x)(333-x)-81^2=0 are x=360,x=90