What are the solutions to the equation (7+x)(9+x)=92 ?

The solutions to the equation (7+x)(9+x)=92 are x=-8+sqrt(93),x=-8-sqrt(93)

Find the zeros of (7+x)(9+x)=92

The zeros of (7+x)(9+x)=92 are x=-8+sqrt(93),x=-8-sqrt(93)

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(7 + x) (9 + x) = 92

x = - 8 + 93, x = - 8 - 93

Decimal Notation

x = 1.64365 \dots, x = - 17.64365 \dots

Solution steps

(7 + x) (9 + x) = 92

Expand

(7 + x) (9 + x) : 63 + 16 x + x^{2}

63 + 16 x + x^{2} = 92

Move

92

to the left side

x^{2} + 16 x - 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot ( - 29 )}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot (- 29) = 293

x_{1, 2} = \frac{- 16 \pm 2 93}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 16 + 2 93}{2 \cdot 1}, x_{2} = \frac{- 16 - 2 93}{2 \cdot 1}

x = \frac{- 16 + 2 93}{2 \cdot 1} : - 8 + 93

x = \frac{- 16 - 2 93}{2 \cdot 1} : - 8 - 93

The solutions to the quadratic equation are:

x = - 8 + 93, x = - 8 - 93

\frac{1}{2} x^{2} + 2 x + 2 = 0

2 k^{2} + k - 7 = 0

10 a^{2} + 5 a = 15

x^{2} - 4 x + 96 = 0

5 x = x^{2} + 1

What are the solutions to the equation (7+x)(9+x)=92 ?
The solutions to the equation (7+x)(9+x)=92 are x=-8+sqrt(93),x=-8-sqrt(93)
Find the zeros of (7+x)(9+x)=92
The zeros of (7+x)(9+x)=92 are x=-8+sqrt(93),x=-8-sqrt(93)