Car factory

Carmaker now produce 2 cars a day more than last year, so the production of 70312 cars will save just one full working day. How many working days needed to manufacture 70312 cars last year?

Result

w =  188 d

Solution:

(x+2) (w1)=70312 wx=70312  (70312/w+2) (w1)=70312  (70312+2w)(w1)=70312w  (70312+2 w) (w1)=70312 w 2w22w70312=0  a=2;b=2;c=70312 D=b24ac=2242(70312)=562500 D>0  w1,2=b±D2a=2±5625004 w1,2=2±7504 w1,2=0.5±187.5 w1=188 w2=187   Factored form of the equation:  2(w188)(w+187)=0  w=w1=188 x1=70312/w=70312/188=374 a/d w=w1=188 d(x+2) \cdot \ (w-1)=70312 \ \\ w x=70312 \ \\ \ \\ (70312/w+2) \cdot \ (w-1)=70312 \ \\ \ \\ (70312+2*w)*(w-1)=70312*w \ \\ \ \\ (70312+2 \cdot \ w) \cdot \ (w-1)=70312 \cdot \ w \ \\ 2w^2 -2w -70312=0 \ \\ \ \\ a=2; b=-2; c=-70312 \ \\ D=b^2 - 4ac=2^2 - 4\cdot 2 \cdot (-70312)=562500 \ \\ D>0 \ \\ \ \\ w_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 2 \pm \sqrt{ 562500 } }{ 4 } \ \\ w_{1,2}=\dfrac{ 2 \pm 750 }{ 4 } \ \\ w_{1,2}=0.5 \pm 187.5 \ \\ w_{1}=188 \ \\ w_{2}=-187 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (w -188) (w +187)=0 \ \\ \ \\ w=w_{1}=188 \ \\ x_{1}=70312/w=70312/188=374 \ \text{a/d} \ \\ w=w_{1}=188 \ \text{d}

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