What are the solutions to the equation x^2-9x=120 ?

The solutions to the equation x^2-9x=120 are x=(9+sqrt(561))/2 ,x=(9-sqrt(561))/2

Find the zeros of x^2-9x=120

The zeros of x^2-9x=120 are x=(9+sqrt(561))/2 ,x=(9-sqrt(561))/2

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x^{2} - 9 x = 120

x = \frac{9 + 561}{2}, x = \frac{9 - 561}{2}

Decimal Notation

x = 16.34271 \dots, x = - 7.34271 \dots

Solution steps

x^{2} - 9 x = 120

Move

120

to the left side

x^{2} - 9 x - 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 120 )}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot (- 120) = 561

x_{1, 2} = \frac{- ( - 9 ) \pm 561}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 561}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 561}{2 \cdot 1}

x = \frac{- ( - 9 ) + 561}{2 \cdot 1} : \frac{9 + 561}{2}

x = \frac{- ( - 9 ) - 561}{2 \cdot 1} : \frac{9 - 561}{2}

The solutions to the quadratic equation are:

x = \frac{9 + 561}{2}, x = \frac{9 - 561}{2}

x^{2} - 65 x + 1050 = 0

6 a^{2} - a - 1 = 0

r, r^{2} - 2 b r + b = 0

(x - 8) (x - 4) = - 4

x, 2 x^{2} + 4 y^{2} + 8 x - 8 y + 12 = 0

What are the solutions to the equation x^2-9x=120 ?
The solutions to the equation x^2-9x=120 are x=(9+sqrt(561))/2 ,x=(9-sqrt(561))/2
Find the zeros of x^2-9x=120
The zeros of x^2-9x=120 are x=(9+sqrt(561))/2 ,x=(9-sqrt(561))/2