What are the solutions to the equation 59=q^2-12q+41 ?

The solutions to the equation 59=q^2-12q+41 are q=3(2+sqrt(6)),q=3(2-sqrt(6))

Find the zeros of 59=q^2-12q+41

The zeros of 59=q^2-12q+41 are q=3(2+sqrt(6)),q=3(2-sqrt(6))

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59 = q^{2} - 12 q + 41

q = 3 (2 + 6), q = 3 (2 - 6)

Solution steps

59 = q^{2} - 12 q + 41

Switch sides

q^{2} - 12 q + 41 = 59

Move

59

to the left side

q^{2} - 12 q - 18 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 18) = 66

q_{1, 2} = \frac{- ( - 12 ) \pm 6 6}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- ( - 12 ) + 6 6}{2 \cdot 1}, q_{2} = \frac{- ( - 12 ) - 6 6}{2 \cdot 1}

q = \frac{- ( - 12 ) + 6 6}{2 \cdot 1} : 3 (2 + 6)

q = \frac{- ( - 12 ) - 6 6}{2 \cdot 1} : 3 (2 - 6)

The solutions to the quadratic equation are:

q = 3 (2 + 6), q = 3 (2 - 6)

3 x^{2} - 130 x + 1000 = 0

2 n^{2} + 5 n + 2 = 0

x (x + 10) + 20 = - 5

10 x^{2} - 60 x = 0

6 x^{2} - 13 x = 5

What are the solutions to the equation 59=q^2-12q+41 ?
The solutions to the equation 59=q^2-12q+41 are q=3(2+sqrt(6)),q=3(2-sqrt(6))
Find the zeros of 59=q^2-12q+41
The zeros of 59=q^2-12q+41 are q=3(2+sqrt(6)),q=3(2-sqrt(6))