What are the solutions to the equation (50-2x)(40-2x)=875 ?

The solutions to the equation (50-2x)(40-2x)=875 are x= 75/2 ,x= 15/2

Find the zeros of (50-2x)(40-2x)=875

The zeros of (50-2x)(40-2x)=875 are x= 75/2 ,x= 15/2

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(50 - 2 x) (40 - 2 x) = 875

x = \frac{75}{2}, x = \frac{15}{2}

Decimal Notation

x = 37.5, x = 7.5

Solution steps

(50 - 2 x) (40 - 2 x) = 875

Expand

(50 - 2 x) (40 - 2 x) : 2000 - 180 x + 4 x^{2}

2000 - 180 x + 4 x^{2} = 875

Move

875

to the left side

4 x^{2} - 180 x + 1125 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 180 ) \pm ( - 180 ) ^{2} - 4 \cdot 4 \cdot 1125}{2 \cdot 4}

(- 180)^{2} - 4 \cdot 4 \cdot 1125 = 120

x_{1, 2} = \frac{- ( - 180 ) \pm 120}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 180 ) + 120}{2 \cdot 4}, x_{2} = \frac{- ( - 180 ) - 120}{2 \cdot 4}

x = \frac{- ( - 180 ) + 120}{2 \cdot 4} : \frac{75}{2}

x = \frac{- ( - 180 ) - 120}{2 \cdot 4} : \frac{15}{2}

The solutions to the quadratic equation are:

x = \frac{75}{2}, x = \frac{15}{2}

n^{2} + 2 n - 2 = 0

x^{12} y^{18} + 1

5 x^{2} - 20 y^{2}

\frac{1}{2} > - 2 x + \frac{7}{2}

∣ 12 ∣

What are the solutions to the equation (50-2x)(40-2x)=875 ?
The solutions to the equation (50-2x)(40-2x)=875 are x= 75/2 ,x= 15/2
Find the zeros of (50-2x)(40-2x)=875
The zeros of (50-2x)(40-2x)=875 are x= 75/2 ,x= 15/2