What are the solutions to the equation j^2-26j+7=0 ?

The solutions to the equation j^2-26j+7=0 are j=13+9sqrt(2),j=13-9sqrt(2)

Find the zeros of j^2-26j+7=0

The zeros of j^2-26j+7=0 are j=13+9sqrt(2),j=13-9sqrt(2)

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j^{2} - 26 j + 7 = 0

j = 13 + 92, j = 13 - 92

Decimal Notation

j = 25.72792 \dots, j = 0.27207 \dots

Solution steps

j^{2} - 26 j + 7 = 0

Solve with the quadratic formula

j_{1, 2} = \frac{- ( - 26 ) \pm ( - 26 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

(- 26)^{2} - 4 \cdot 1 \cdot 7 = 182

j_{1, 2} = \frac{- ( - 26 ) \pm 18 2}{2 \cdot 1}

Separate the solutions

j_{1} = \frac{- ( - 26 ) + 18 2}{2 \cdot 1}, j_{2} = \frac{- ( - 26 ) - 18 2}{2 \cdot 1}

j = \frac{- ( - 26 ) + 18 2}{2 \cdot 1} : 13 + 92

j = \frac{- ( - 26 ) - 18 2}{2 \cdot 1} : 13 - 92

The solutions to the quadratic equation are:

j = 13 + 92, j = 13 - 92

\frac{u ^{2 x + 1}}{u ^{x}}

∣ 3 - 7 x ∣ = - 6

- 3 + 6 i - (- 5 - 3 i) - 8 i

\frac{( 1 + 2 i )}{( 2 + 3 i )}

2 (x + 5) (x - 6) = 0

What are the solutions to the equation j^2-26j+7=0 ?
The solutions to the equation j^2-26j+7=0 are j=13+9sqrt(2),j=13-9sqrt(2)
Find the zeros of j^2-26j+7=0
The zeros of j^2-26j+7=0 are j=13+9sqrt(2),j=13-9sqrt(2)