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?

Correct 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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