What are the solutions to the equation (x+4)(x+3)=9^2 ?

The solutions to the equation (x+4)(x+3)=9^2 are x=(-7+5sqrt(13))/2 ,x=(-7-5sqrt(13))/2

Find the zeros of (x+4)(x+3)=9^2

The zeros of (x+4)(x+3)=9^2 are x=(-7+5sqrt(13))/2 ,x=(-7-5sqrt(13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x + 4) (x + 3) = 9^{2}

x = \frac{- 7 + 5 13}{2}, x = \frac{- 7 - 5 13}{2}

Decimal Notation

x = 5.51387 \dots, x = - 12.51387 \dots

Solution steps

(x + 4) (x + 3) = 9^{2}

Expand

(x + 4) (x + 3) : x^{2} + 7 x + 12

Expand

9^{2} : 81

x^{2} + 7 x + 12 = 81

Move

81

to the left side

x^{2} + 7 x - 69 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 69 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 69) = 513

x_{1, 2} = \frac{- 7 \pm 5 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 5 13}{2 \cdot 1}, x_{2} = \frac{- 7 - 5 13}{2 \cdot 1}

x = \frac{- 7 + 5 13}{2 \cdot 1} : \frac{- 7 + 5 13}{2}

x = \frac{- 7 - 5 13}{2 \cdot 1} : \frac{- 7 - 5 13}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 5 13}{2}, x = \frac{- 7 - 5 13}{2}

x^{2} - 7 x + 14 = 7

18 x^{2} - 2 = 0

v, h = \frac{v ^{2}}{2 g}

x, y^{3} - x^{2} + 3 y^{2} + 2 x + 3 y = 0

f^{2} = 25

What are the solutions to the equation (x+4)(x+3)=9^2 ?
The solutions to the equation (x+4)(x+3)=9^2 are x=(-7+5sqrt(13))/2 ,x=(-7-5sqrt(13))/2
Find the zeros of (x+4)(x+3)=9^2
The zeros of (x+4)(x+3)=9^2 are x=(-7+5sqrt(13))/2 ,x=(-7-5sqrt(13))/2