What are the solutions to the equation (20+x)(35+x)=1270 ?

The solutions to the equation (20+x)(35+x)=1270 are x=(-55+sqrt(5305))/2 ,x=(-55-sqrt(5305))/2

Find the zeros of (20+x)(35+x)=1270

The zeros of (20+x)(35+x)=1270 are x=(-55+sqrt(5305))/2 ,x=(-55-sqrt(5305))/2

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(20 + x) (35 + x) = 1270

x = \frac{- 55 + 5305}{2}, x = \frac{- 55 - 5305}{2}

Decimal Notation

x = 8.91771 \dots, x = - 63.91771 \dots

Solution steps

(20 + x) (35 + x) = 1270

Expand

(20 + x) (35 + x) : 700 + 55 x + x^{2}

700 + 55 x + x^{2} = 1270

Move

1270

to the left side

x^{2} + 55 x - 570 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 55 \pm 5 5 ^{2} - 4 \cdot 1 \cdot ( - 570 )}{2 \cdot 1}

5 5^{2} - 4 \cdot 1 \cdot (- 570) = 5305

x_{1, 2} = \frac{- 55 \pm 5305}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 55 + 5305}{2 \cdot 1}, x_{2} = \frac{- 55 - 5305}{2 \cdot 1}

x = \frac{- 55 + 5305}{2 \cdot 1} : \frac{- 55 + 5305}{2}

x = \frac{- 55 - 5305}{2 \cdot 1} : \frac{- 55 - 5305}{2}

The solutions to the quadratic equation are:

x = \frac{- 55 + 5305}{2}, x = \frac{- 55 - 5305}{2}

2 x (5 x - 1) = 9

10 x^{2} - 16 x - 8 = 0

0.6 x^{2} + 6 x + 2 = 0

x^{2} - x - 20

9 u^{2} - 30 u = - 25

What are the solutions to the equation (20+x)(35+x)=1270 ?
The solutions to the equation (20+x)(35+x)=1270 are x=(-55+sqrt(5305))/2 ,x=(-55-sqrt(5305))/2
Find the zeros of (20+x)(35+x)=1270
The zeros of (20+x)(35+x)=1270 are x=(-55+sqrt(5305))/2 ,x=(-55-sqrt(5305))/2