What are the solutions to the equation (30+x)(35+x)=1300 ?

The solutions to the equation (30+x)(35+x)=1300 are x=(-65+5sqrt(209))/2 ,x=(-65-5sqrt(209))/2

Find the zeros of (30+x)(35+x)=1300

The zeros of (30+x)(35+x)=1300 are x=(-65+5sqrt(209))/2 ,x=(-65-5sqrt(209))/2

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(30 + x) (35 + x) = 1300

x = \frac{- 65 + 5 209}{2}, x = \frac{- 65 - 5 209}{2}

Decimal Notation

x = 3.64208 \dots, x = - 68.64208 \dots

Solution steps

(30 + x) (35 + x) = 1300

Expand

(30 + x) (35 + x) : 1050 + 65 x + x^{2}

1050 + 65 x + x^{2} = 1300

Move

1300

to the left side

x^{2} + 65 x - 250 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 65 \pm 6 5 ^{2} - 4 \cdot 1 \cdot ( - 250 )}{2 \cdot 1}

6 5^{2} - 4 \cdot 1 \cdot (- 250) = 5209

x_{1, 2} = \frac{- 65 \pm 5 209}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 65 + 5 209}{2 \cdot 1}, x_{2} = \frac{- 65 - 5 209}{2 \cdot 1}

x = \frac{- 65 + 5 209}{2 \cdot 1} : \frac{- 65 + 5 209}{2}

x = \frac{- 65 - 5 209}{2 \cdot 1} : \frac{- 65 - 5 209}{2}

The solutions to the quadratic equation are:

x = \frac{- 65 + 5 209}{2}, x = \frac{- 65 - 5 209}{2}

0 = - x^{2} + 42 x - 20

m^{2} - m + 3 = 0

0 = - x^{2} - 3 x

x^{2} + 10 x = - 10

\frac{1}{2} n^{2} + \frac{2}{3} n = - 1

What are the solutions to the equation (30+x)(35+x)=1300 ?
The solutions to the equation (30+x)(35+x)=1300 are x=(-65+5sqrt(209))/2 ,x=(-65-5sqrt(209))/2
Find the zeros of (30+x)(35+x)=1300
The zeros of (30+x)(35+x)=1300 are x=(-65+5sqrt(209))/2 ,x=(-65-5sqrt(209))/2