What are the solutions to the equation (35-x)(26-x)=850 ?

The solutions to the equation (35-x)(26-x)=850 are x=60,x=1

Find the zeros of (35-x)(26-x)=850

The zeros of (35-x)(26-x)=850 are x=60,x=1

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(35 - x) (26 - x) = 850

x = 60, x = 1

Solution steps

(35 - x) (26 - x) = 850

Expand

(35 - x) (26 - x) : 910 - 61 x + x^{2}

910 - 61 x + x^{2} = 850

Move

850

to the left side

x^{2} - 61 x + 60 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 61 ) \pm ( - 61 ) ^{2} - 4 \cdot 1 \cdot 60}{2 \cdot 1}

(- 61)^{2} - 4 \cdot 1 \cdot 60 = 59

x_{1, 2} = \frac{- ( - 61 ) \pm 59}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 61 ) + 59}{2 \cdot 1}, x_{2} = \frac{- ( - 61 ) - 59}{2 \cdot 1}

x = \frac{- ( - 61 ) + 59}{2 \cdot 1} : 60

x = \frac{- ( - 61 ) - 59}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 60, x = 1

(2 p + 3) (p - 5) = 0

(z + 6)^{2} = 64

45 m^{2} + 0 + 4 = 0

- 4 x^{2} + 30 x + 54 = 0

34 x^{2} + 100 x - 125 = 0

What are the solutions to the equation (35-x)(26-x)=850 ?
The solutions to the equation (35-x)(26-x)=850 are x=60,x=1
Find the zeros of (35-x)(26-x)=850
The zeros of (35-x)(26-x)=850 are x=60,x=1