What are the solutions to the equation (20+x)(30+x)=1690 ?

The solutions to the equation (20+x)(30+x)=1690 are x=-25+7sqrt(35),x=-25-7sqrt(35)

Find the zeros of (20+x)(30+x)=1690

The zeros of (20+x)(30+x)=1690 are x=-25+7sqrt(35),x=-25-7sqrt(35)

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(20 + x) (30 + x) = 1690

x = - 25 + 735, x = - 25 - 735

Decimal Notation

x = 16.41255 \dots, x = - 66.41255 \dots

Solution steps

(20 + x) (30 + x) = 1690

Expand

(20 + x) (30 + x) : 600 + 50 x + x^{2}

600 + 50 x + x^{2} = 1690

Move

1690

to the left side

x^{2} + 50 x - 1090 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 \cdot 1 \cdot ( - 1090 )}{2 \cdot 1}

5 0^{2} - 4 \cdot 1 \cdot (- 1090) = 1435

x_{1, 2} = \frac{- 50 \pm 14 35}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 50 + 14 35}{2 \cdot 1}, x_{2} = \frac{- 50 - 14 35}{2 \cdot 1}

x = \frac{- 50 + 14 35}{2 \cdot 1} : - 25 + 735

x = \frac{- 50 - 14 35}{2 \cdot 1} : - 25 - 735

The solutions to the quadratic equation are:

x = - 25 + 735, x = - 25 - 735

x^{2} + 3 x + 3 = 10

x (x + 2)^{2} = (x + 1)^{3} + 5

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

x, (x - 1)^{2} + (y - 2)^{2} = 0

A = x^{2} + 6 x + 5

What are the solutions to the equation (20+x)(30+x)=1690 ?
The solutions to the equation (20+x)(30+x)=1690 are x=-25+7sqrt(35),x=-25-7sqrt(35)
Find the zeros of (20+x)(30+x)=1690
The zeros of (20+x)(30+x)=1690 are x=-25+7sqrt(35),x=-25-7sqrt(35)