What are the solutions to the equation (18+2x)(26+2x)=1140 ?

The solutions to the equation (18+2x)(26+2x)=1140 are x=6,x=-28

Find the zeros of (18+2x)(26+2x)=1140

The zeros of (18+2x)(26+2x)=1140 are x=6,x=-28

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(18 + 2 x) (26 + 2 x) = 1140

x = 6, x = - 28

Solution steps

(18 + 2 x) (26 + 2 x) = 1140

Expand

(18 + 2 x) (26 + 2 x) : 468 + 88 x + 4 x^{2}

468 + 88 x + 4 x^{2} = 1140

Move

1140

to the left side

4 x^{2} + 88 x - 672 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 88 \pm 8 8 ^{2} - 4 \cdot 4 ( - 672 )}{2 \cdot 4}

8 8^{2} - 4 \cdot 4 (- 672) = 136

x_{1, 2} = \frac{- 88 \pm 136}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 88 + 136}{2 \cdot 4}, x_{2} = \frac{- 88 - 136}{2 \cdot 4}

x = \frac{- 88 + 136}{2 \cdot 4} : 6

x = \frac{- 88 - 136}{2 \cdot 4} : - 28

The solutions to the quadratic equation are:

x = 6, x = - 28

- 4 (2 x + 7) = 2 x - 18

9 x \leq 12 x^{2}

1 < 1 - \frac{1}{2} x < 3

\frac{2}{3} - \frac{2}{7}

y^{2} + y - 30

What are the solutions to the equation (18+2x)(26+2x)=1140 ?
The solutions to the equation (18+2x)(26+2x)=1140 are x=6,x=-28
Find the zeros of (18+2x)(26+2x)=1140
The zeros of (18+2x)(26+2x)=1140 are x=6,x=-28