What are the solutions to the equation x^2-8x-5=7 ?

The solutions to the equation x^2-8x-5=7 are x=2(2+sqrt(7)),x=2(2-sqrt(7))

Find the zeros of x^2-8x-5=7

The zeros of x^2-8x-5=7 are x=2(2+sqrt(7)),x=2(2-sqrt(7))

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x^{2} - 8 x - 5 = 7

x = 2 (2 + 7), x = 2 (2 - 7)

Decimal Notation

x = 9.29150 \dots, x = - 1.29150 \dots

Solution steps

x^{2} - 8 x - 5 = 7

Move

7

to the left side

x^{2} - 8 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 12) = 47

x_{1, 2} = \frac{- ( - 8 ) \pm 4 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 7}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 4 7}{2 \cdot 1}

x = \frac{- ( - 8 ) + 4 7}{2 \cdot 1} : 2 (2 + 7)

x = \frac{- ( - 8 ) - 4 7}{2 \cdot 1} : 2 (2 - 7)

The solutions to the quadratic equation are:

x = 2 (2 + 7), x = 2 (2 - 7)

2 w^{2} - 5 w - 14 = - 5

3 x^{2} + 6 x + 1

2 + x - 6 x^{2} = 0

\frac{t ^{2} - 1}{3} = t^{2} - 51

a, 9 a = 16 - a^{2}

What are the solutions to the equation x^2-8x-5=7 ?
The solutions to the equation x^2-8x-5=7 are x=2(2+sqrt(7)),x=2(2-sqrt(7))
Find the zeros of x^2-8x-5=7
The zeros of x^2-8x-5=7 are x=2(2+sqrt(7)),x=2(2-sqrt(7))