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2/5(h)-7=12/5(h)-2h+3
We move all terms to the left:
2/5(h)-7-(12/5(h)-2h+3)=0
Domain of the equation: 5h!=0
h!=0/5
h!=0
h∈R
Domain of the equation: 5h-2h+3)!=0We add all the numbers together, and all the variables
h∈R
2/5h-(-2h+12/5h+3)-7=0
We get rid of parentheses
2/5h+2h-12/5h-3-7=0
We multiply all the terms by the denominator
2h*5h-3*5h-7*5h+2-12=0
We add all the numbers together, and all the variables
2h*5h-3*5h-7*5h-10=0
Wy multiply elements
10h^2-15h-35h-10=0
We add all the numbers together, and all the variables
10h^2-50h-10=0
a = 10; b = -50; c = -10;
Δ = b2-4ac
Δ = -502-4·10·(-10)
Δ = 2900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2900}=\sqrt{100*29}=\sqrt{100}*\sqrt{29}=10\sqrt{29}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{29}}{2*10}=\frac{50-10\sqrt{29}}{20} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{29}}{2*10}=\frac{50+10\sqrt{29}}{20} $
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