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0.4x+10=3/5x+12=
We move all terms to the left:
0.4x+10-(3/5x+12)=0
Domain of the equation: 5x+12)!=0We get rid of parentheses
x∈R
0.4x-3/5x-12+10=0
We multiply all the terms by the denominator
(0.4x)*5x-12*5x+10*5x-3=0
We add all the numbers together, and all the variables
(+0.4x)*5x-12*5x+10*5x-3=0
We multiply parentheses
0x^2-12*5x+10*5x-3=0
Wy multiply elements
0x^2-60x+50x-3=0
We add all the numbers together, and all the variables
x^2-10x-3=0
a = 1; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·1·(-3)
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{7}}{2*1}=\frac{10-4\sqrt{7}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{7}}{2*1}=\frac{10+4\sqrt{7}}{2} $
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