What are the solutions to the equation (x-3)(x-5)=35 ?

The solutions to the equation (x-3)(x-5)=35 are x=10,x=-2

Find the zeros of (x-3)(x-5)=35

The zeros of (x-3)(x-5)=35 are x=10,x=-2

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(x - 3) (x - 5) = 35

x = 10, x = - 2

Solution steps

(x - 3) (x - 5) = 35

Expand

(x - 3) (x - 5) : x^{2} - 8 x + 15

x^{2} - 8 x + 15 = 35

Move

35

to the left side

x^{2} - 8 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 20) = 12

x_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 12}{2 \cdot 1}

x = \frac{- ( - 8 ) + 12}{2 \cdot 1} : 10

x = \frac{- ( - 8 ) - 12}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 10, x = - 2

s^{2} + 5 s + 4

24 x - 8 = - 18 x + 13

(5 x - 3) (x - 5) < 0

f = 3^{2}

- 10 + 3 x = 8

What are the solutions to the equation (x-3)(x-5)=35 ?
The solutions to the equation (x-3)(x-5)=35 are x=10,x=-2
Find the zeros of (x-3)(x-5)=35
The zeros of (x-3)(x-5)=35 are x=10,x=-2