What are the solutions to the equation x^2-12x-26=0 ?

The solutions to the equation x^2-12x-26=0 are x=6+sqrt(62),x=6-sqrt(62)

Find the zeros of x^2-12x-26=0

The zeros of x^2-12x-26=0 are x=6+sqrt(62),x=6-sqrt(62)

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x^{2} - 12 x - 26 = 0

x = 6 + 62, x = 6 - 62

Solution steps

x^{2} - 12 x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 26 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 26) = 262

x_{1, 2} = \frac{- ( - 12 ) \pm 2 62}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 62}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 2 62}{2 \cdot 1}

x = \frac{- ( - 12 ) + 2 62}{2 \cdot 1} : 6 + 62

x = \frac{- ( - 12 ) - 2 62}{2 \cdot 1} : 6 - 62

The solutions to the quadratic equation are:

x = 6 + 62, x = 6 - 62

x^{2} - 1 = lo g_{3} (27)

4 x^{2} + 18 x + 18 = 0

x^{2} + 18 x + 81 = 49

x^{2} - 1 = 6 x

3 x - 2 x^{2} - 7 = 0

What are the solutions to the equation x^2-12x-26=0 ?
The solutions to the equation x^2-12x-26=0 are x=6+sqrt(62),x=6-sqrt(62)
Find the zeros of x^2-12x-26=0
The zeros of x^2-12x-26=0 are x=6+sqrt(62),x=6-sqrt(62)