What are the solutions to the equation x^2+16x=41 ?

The solutions to the equation x^2+16x=41 are x=-8+sqrt(105),x=-8-sqrt(105)

Find the zeros of x^2+16x=41

The zeros of x^2+16x=41 are x=-8+sqrt(105),x=-8-sqrt(105)

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x^{2} + 16 x = 41

x = - 8 + 105, x = - 8 - 105

Decimal Notation

x = 2.24695 \dots, x = - 18.24695 \dots

Solution steps

x^{2} + 16 x = 41

Move

41

to the left side

x^{2} + 16 x - 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot ( - 41 )}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot (- 41) = 2105

x_{1, 2} = \frac{- 16 \pm 2 105}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 16 + 2 105}{2 \cdot 1}, x_{2} = \frac{- 16 - 2 105}{2 \cdot 1}

x = \frac{- 16 + 2 105}{2 \cdot 1} : - 8 + 105

x = \frac{- 16 - 2 105}{2 \cdot 1} : - 8 - 105

The solutions to the quadratic equation are:

x = - 8 + 105, x = - 8 - 105

(x - 2)^{2} + 2 = x

\frac{1}{3} π r^{2} (3) = 25 π

x, 4 (y - 2 x + 3) = (x + c)^{2}

x^{2} + (- \frac{2}{2})^{2} = 1

(2 p + 3)^{2} = p^{2} - 5 p + 9

What are the solutions to the equation x^2+16x=41 ?
The solutions to the equation x^2+16x=41 are x=-8+sqrt(105),x=-8-sqrt(105)
Find the zeros of x^2+16x=41
The zeros of x^2+16x=41 are x=-8+sqrt(105),x=-8-sqrt(105)