What are the solutions to the equation (14+2x)(26+2x)=1408 ?

The solutions to the equation (14+2x)(26+2x)=1408 are x=9,x=-29

Find the zeros of (14+2x)(26+2x)=1408

The zeros of (14+2x)(26+2x)=1408 are x=9,x=-29

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(14 + 2 x) (26 + 2 x) = 1408

x = 9, x = - 29

Solution steps

(14 + 2 x) (26 + 2 x) = 1408

Expand

(14 + 2 x) (26 + 2 x) : 364 + 80 x + 4 x^{2}

364 + 80 x + 4 x^{2} = 1408

Move

1408

to the left side

4 x^{2} + 80 x - 1044 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 4 ( - 1044 )}{2 \cdot 4}

8 0^{2} - 4 \cdot 4 (- 1044) = 152

x_{1, 2} = \frac{- 80 \pm 152}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 80 + 152}{2 \cdot 4}, x_{2} = \frac{- 80 - 152}{2 \cdot 4}

x = \frac{- 80 + 152}{2 \cdot 4} : 9

x = \frac{- 80 - 152}{2 \cdot 4} : - 29

The solutions to the quadratic equation are:

x = 9, x = - 29

x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}

3 d^{2} + d = - 2 d

- 1 < 2 x - 5 < 7

(0.5 a - 0.6 b + 5.5) - (- 0.5 a + 0.4 b) + (1.3 b - 4.5)

x^{2} + 6 x = 14

What are the solutions to the equation (14+2x)(26+2x)=1408 ?
The solutions to the equation (14+2x)(26+2x)=1408 are x=9,x=-29
Find the zeros of (14+2x)(26+2x)=1408
The zeros of (14+2x)(26+2x)=1408 are x=9,x=-29