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10x+19x+6x^2=0
We add all the numbers together, and all the variables
6x^2+29x=0
a = 6; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·6·0
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*6}=\frac{-58}{12} =-4+5/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*6}=\frac{0}{12} =0 $
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