What are the solutions to the equation q^2-22q+120=0 ?

The solutions to the equation q^2-22q+120=0 are q=12,q=10

Find the zeros of q^2-22q+120=0

The zeros of q^2-22q+120=0 are q=12,q=10

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q^{2} - 22 q + 120 = 0

q = 12, q = 10

Solution steps

q^{2} - 22 q + 120 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 120}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 120 = 2

q_{1, 2} = \frac{- ( - 22 ) \pm 2}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- ( - 22 ) + 2}{2 \cdot 1}, q_{2} = \frac{- ( - 22 ) - 2}{2 \cdot 1}

q = \frac{- ( - 22 ) + 2}{2 \cdot 1} : 12

q = \frac{- ( - 22 ) - 2}{2 \cdot 1} : 10

The solutions to the quadratic equation are:

q = 12, q = 10

(x - 1) (x - 2) = 2

x, c y^{2} = x^{2} + y

(2 - 3 x)^{2} = 0

5 x^{2} + 6 x = - 1

(x + 9) (x - 2) = 60

What are the solutions to the equation q^2-22q+120=0 ?
The solutions to the equation q^2-22q+120=0 are q=12,q=10
Find the zeros of q^2-22q+120=0
The zeros of q^2-22q+120=0 are q=12,q=10