What are the solutions to the equation x^2-15x+36=0 ?

The solutions to the equation x^2-15x+36=0 are x=12,x=3

Find the zeros of x^2-15x+36=0

The zeros of x^2-15x+36=0 are x=12,x=3

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x^{2} - 15 x + 36 = 0

x = 12, x = 3

Solution steps

x^{2} - 15 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 36 = 9

x_{1, 2} = \frac{- ( - 15 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 15 ) - 9}{2 \cdot 1}

x = \frac{- ( - 15 ) + 9}{2 \cdot 1} : 12

x = \frac{- ( - 15 ) - 9}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

x = 12, x = 3

- \frac{1}{2} + \frac{5}{4}

6 x - 35 + 3 x + 53 = 180

(t^{- 4})^{- 9} t^{2}

5 (2 x + 4) = 40

5 y^{2} - 15 y

What are the solutions to the equation x^2-15x+36=0 ?
The solutions to the equation x^2-15x+36=0 are x=12,x=3
Find the zeros of x^2-15x+36=0
The zeros of x^2-15x+36=0 are x=12,x=3