What are the solutions to the equation (40-x)(60-x)=800 ?

The solutions to the equation (40-x)(60-x)=800 are x=80,x=20

Find the zeros of (40-x)(60-x)=800

The zeros of (40-x)(60-x)=800 are x=80,x=20

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(40 - x) (60 - x) = 800

x = 80, x = 20

Solution steps

(40 - x) (60 - x) = 800

Expand

(40 - x) (60 - x) : 2400 - 100 x + x^{2}

2400 - 100 x + x^{2} = 800

Move

800

to the left side

x^{2} - 100 x + 1600 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 100 ) \pm ( - 100 ) ^{2} - 4 \cdot 1 \cdot 1600}{2 \cdot 1}

(- 100)^{2} - 4 \cdot 1 \cdot 1600 = 60

x_{1, 2} = \frac{- ( - 100 ) \pm 60}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 100 ) + 60}{2 \cdot 1}, x_{2} = \frac{- ( - 100 ) - 60}{2 \cdot 1}

x = \frac{- ( - 100 ) + 60}{2 \cdot 1} : 80

x = \frac{- ( - 100 ) - 60}{2 \cdot 1} : 20

The solutions to the quadratic equation are:

x = 80, x = 20

2 x^{2} + 160 = 36 x

y, y^{2} + 2 y + 1 = 0

t, A t - \frac{( B t ^{2} )}{2} = 0

(2 x - 1) (x + 2) = 8 x - 4

3 x^{2} - 26 = 1

What are the solutions to the equation (40-x)(60-x)=800 ?
The solutions to the equation (40-x)(60-x)=800 are x=80,x=20
Find the zeros of (40-x)(60-x)=800
The zeros of (40-x)(60-x)=800 are x=80,x=20