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10x^2+2x=18x
We move all terms to the left:
10x^2+2x-(18x)=0
We add all the numbers together, and all the variables
10x^2-16x=0
a = 10; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·10·0
Δ = 256
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*10}=\frac{0}{20} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*10}=\frac{32}{20} =1+3/5 $
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